Question 13 Marks
The general term of a sequence is give by $a_n = -4n + 15.$ Is the sequence an A.P.?
If so, find its $15^{th}$ term and the common difference.
AnswerGeneral term of a sequence
$a_n = -4n + 15$
Let $n = 1, 2, 3, 4, 5, .....,$ then
$a_1 = -4 \times 1 + 15 = -4 + 15 = 11$
$a_2 = -4 \times 2 + 15 = -8 + 15 = 7$
$a_3 = -4 \times 3 + 15 = -12 + 15 = 3$
$a_4 = -4 \times 4 + 15 = -16 + 15 = -1$
$a_5 = -4 \times 5 + 15 = -20 + 15 = -5$
We see that first term is $11$ and common difference is $-4$
$a_2 - a_1 = 7 - 11 = -4$
$a_3 - a_2 = 3 - 7 = -4$
$a_4 - a_3 = -1 - 3 = -4$
$a_5 - a_4 = -5 - (-1) = -5 + 1 = -4$
$\therefore$ Yes, it is an A.P.
Now $15^{th}$ term $= a_{15} = -4 \times 15 + 15$
$= -60 + 15 = -45.$
View full question & answer→Question 23 Marks
Find:$9^{th}$ term of the A.P. $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}, .....$
AnswerGiven A.P. is
$\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}, .....$
First term $(\text{a})=\frac{3}{4}$
Common difference (d) = Second - First term
$=\frac{5}{4}-\frac{3}{4}$
$=\frac{2}{4}$
$n^{th}$ term $a_n = a + (n - 1)d$
$9^{th}$ term $a_9 = a + (9 - 1)d$
$=\frac{3}{4}+8.\frac{2}{4}$
$=\frac{3}{4}+\frac{16}{4}$
$=\frac{19}{4}$
View full question & answer→Question 33 Marks
Sum of 13 terms of the A.P. -6, 0, 6, 12, .....
AnswerGiven,
A.P. is -6, 0, 6, 12, .....
Here,
First term a = -6
Difference, d = 0 - (-6) = 6
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{13}=\frac{13}{2}[2(-6)+(13-1)6]$
$\Rightarrow\ \text{S}_{13}=\frac{13}{2}[-12+12\times6]$
$\Rightarrow\ \text{S}_{13}=\frac{13}{2}\times60$
$\Rightarrow\ \text{S}_{13}=390$
Hence, Sum of 13 terms is 390.
View full question & answer→Question 43 Marks
Find the sum of the first $15$ terms of each of the following sequences having $n^{th}$ term as:
$b_n = 5 + 2n.$
AnswerGiven,
$b_n = 5 + 2n$
Put $n = 1, b_1 = 5 + 2(1) = 7$
Put $n = 15, b_{15} = 5 + 2(15) = 35 = l$
Sum of $15$ terms $\text{S}_{15}=\frac{15}{2}(7+35)\ \Big(\therefore\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})\Big)$
$=\frac{15}{2}\times42$
$=315$
$\therefore\text{S}_{15}=315$
View full question & answer→Question 53 Marks
Find the sum of the first $15$ terms of each of the following sequences having $n^{th}$ term as:
$y_n = 9 - 5n.$
Answer$y_n = 9 - 5n$ and number of terms $= 15$
$y_1 = 9 - 5 \times 1 = 9 - 5 = 4$
$y_2 = 9 - 5 \times 2 = 9 - 10 = -1$
$\therefore$ First term $(a) = 4$
Common dofference $(d) = y_2 - y_1 = -1 - 4 = -5$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{15}=\frac{15}{2}[2\text{a}+(15 - 1)\text{d}]$
$=\frac{15}{2}[2\times4+(15 - 1)(-5)]$
$=\frac{15}{2}[8+14(-5)]=\frac{15}{2}[8-70]$
$=\frac{15}{2}(-62)=15\times(-31)=-465$
View full question & answer→Question 63 Marks
Find the indicated terms in the following sequences whose $n^{th}$ terms are:
$a_n = (-1)^nn; a_3, a_5, a_8.$
Answer$a_n = (-1)^nn.$
We need to find $a_3, a_5$_ and $a_8$
Now, to find $a_3$_ term we use $n = 3$, we get,
$a_3 = (-1)^3 3$
$= (-1) 3$
$= -3$
$$Also, to find a5 term we use $n = 5,$ we get,
$a_5 = (-1)^5 5$
$= (-1) 5$
$= -5$
Similarly, to find $a_8$ term we use $n = 8$, we get,
$a_8 = (-1)^8 8$
$= (1) 8$
$= 8$
Thus, $a_3 = -3, a_5 = -5$ and $a_8 = 8.$
View full question & answer→Question 73 Marks
Find the common difference of the A.P. and write the next two terms:
$119, 136, 153, 170, .....$
Answer$119, 136, 153, 170, .....$
Here,
$a_1 = 119$
$a_2 = 136$
So, common difference of the A.P. $(d) = a_2 - a_1$
$= 136 - 119$
$= 17$
Also, we need to find the next two terms of A.P., which means we have to find the $5^{th}$ and $6^{th}$ terms.
So, for fifth term,
$a_5 = a_1 + 4d$
$= 119 + 4(17)$
$= 119 + 68$
$= 187$
Similarly, we find the sixth term,
$a_6 = a_1 + 5d$
$= 119 + 5(17)$
$= 119 + 85$
$= 204$
Therefore, the common difference is $d = 17$ and the next two terms of the A.P. are $a_5 = 187, a_6 = 204.$
View full question & answer→Question 83 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{\text{n}(\text{n}-2)}{2}$
Answer$\text{a}_\text{n}=\frac{\text{n}(\text{n}-2)}{2}$
Let $n = 1, 2, 3, 4, 5,$ then
$\text{a}_1=\frac{1(1-2)}{2}=\frac{1\times(-1)}{2}=\frac{-1}{2}$
$\text{a}_2=\frac{1(2-2)}{2}=\frac{2\times0}{2}=0$
$\text{a}_3=\frac{1(3-2)}{2}=\frac{3\times1}{2}=\frac{3}{2}$
$\text{a}_4=\frac{4(4-2)}{2}=\frac{4\times2}{2}=4$
$\text{a}_5=\frac{5(5-2)}{2}=\frac{5\times3}{2}=\frac{15}{2}$
View full question & answer→Question 93 Marks
The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.
AnswerLet the four consecutive numbers in A.P. be
a - 3d, a - d, a + d, a + 3d
So, a - 3d + a - d + a + d + a + 3d = 32
or 4a = 32
or a = 8
Also, $\frac{(\text{a}-3\text{d})(\text{a}+3\text{d})}{(\text{a}-\text{d})(\text{a}+\text{d})}=\frac{7}{15}$
$\frac{\text{a}^2-9\text{d}^2}{\text{a}^2-\text{d}^2}=\frac{7}{15}$
$15\text{a}^2-135\text{d}^2=7\text{a}^2-7\text{d}^2$
$8\text{a}^2-128\text{d}^2=0$
$\text{d}^2=\frac{8\times8\times8}{128}=4$
$\text{d}=\pm2$
So, when a = 8, d = 2
The numbers are 2, 6, 10, 14.
View full question & answer→Question 103 Marks
Two arithmetic progression have the same common difference. The difference between their $100^{\text {th }}$ terms is $100$ , What is the difference between their $1000^{\text {th }}$ terms?
AnswerLet the two A.P. is be $a_1, a_2, a_3, \ldots .$. and $b_1, b_2, b 3, \ldots .$.
$a_n=a_1+(n-1) d \text { and } b_n=b_1+(n-1) d$
Since common difference of two equations is same given difference between $100^{\text {th }}$ terms is $100$
$a_{100}-b_{100}=100$
$a_1+(99) d-b_1-99 d=100$
$a_1-b_1=100 \ldots . . .(i)$
Difference between. $1000th$ terms is
$a_{1000}-b_{1000}=a_1+(1000-1) d-\left(b_1+(1000-1) d\right)$
$=a_1+999 d-b_1-999 d$
$=a_1-b_1$
$=100(\text { from }(1))$
$\therefore$ Hence difference between $1000^{\text {th }}$ terms of two A.P. is $100 .$
View full question & answer→Question 113 Marks
In a certain A.P. the $24^{th}$ term is twice the $10^{th}$ term. Prove that the $72^{nd}$ term is twice the $34^{th}$ term.
AnswerGiven,
$24^{\text {th }}$ term is twice the $10^{\text {th }}$ term
$a_{24}=2 a_{10}$
Let, first term of a square $=a$
Common difference $= d$
$n^{\text {th }} \text { term } a_n=a+(n-1) d$
$a+(24-1) d=(a+(10-1) d) 2$
$a+23 d=2(a+9 d)$
$(23-18) d=a$
$a=5 d$
We have to prove
$72^{\text {nd }}$ term is twise the $34^{\text {th }}$ term
$a_{72}=2 a_{34}$
$a+(72-1) d=2[a+(34-1) d]$
$a+71 d=2 a=66 d$
Substitute $a=5 d$
$5 d+71 d=2(5 d)+66 d$
$76 d=10 d+66 d$
$76 d=76 d$
Hence proved.
View full question & answer→Question 123 Marks
If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its $(63)^{rd}$ term.
Answer
$-2\text{d}=\frac{7-9}{63}$
$-2\text{d}=\frac{-2}{63}\ \therefore\ \text{d}=\frac{1}{63}$
$\text{a}+6\Big(\frac{1}{63}\Big)=\frac{1}{9}$
$\text{a}=\frac{1}{9}-\frac{6}{63}=\frac{7-6}{63}=\frac{1}{63}$
$\text{a}_{63}=\text{a}+62\text{d}$
$=\frac{1}{63}+62\Big(\frac{1}{63}\Big)=\frac{1+ 62}{63}=\frac{63}{63}=1$ View full question & answer→Question 133 Marks
Find:
Is $-150$ a term of the A.P. $11,8,5,2, \ldots . . ?$
AnswerIn the given problem, we are given an A.P. and the Value of one of its term.
We need to find whether it is a term of the A.P. or not so here we will use the formula $a_n=a+(n-1) d$.
Here,
A.P. is $11,8,5,2, \ldots .$.
$a_n=-150, a=11$ and $d=8-11=-3$
Thus, using the above mentioned formula, we get
$-150=11+(n-1)(-3)$
$\Rightarrow-150-11=-3 n+3$
$\Rightarrow-161=-3 n+3$
$\Rightarrow-161-3=-3 n$
$\Rightarrow-3 n=-164$
$\Rightarrow n=\frac{164}{3}$
Since, the value of n is a fraction. Thus, $-150$ is not the term of the given A.P.
View full question & answer→Question 143 Marks
Find the sum of the first $15$ terms of each of the following sequences having $n^{th}$ term as:
$x_n = 6 - n.$
AnswerHere, we are given an A.P. whose $n^{th}$ term is given bt the following expression, $x_n = 6 - x$, We need to find the sum of first $15$ terms.
So, here we can find the sum of the n term of the given A.P., using the formula,
$\text{S}_\text{n}=\Big(\frac{\text{n}}{2}\Big)(\text{a}+\text{l})$
Where, a = the first term
l = the last term
So, for the given A.P,
The firest term (a) will be calculated using $n = 1$ in the given equation for $n^{th}$ term of A.p.
$x = 6 - 1$
$= 5$
Now, the last term (l) or the $n^{th}$ term is given
$l = a_{15} = 6 - 15$
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
$\text{S}_{15}=\Big(\frac{15}{2}\Big)[(5)+6-15]$
$=\Big(\frac{15}{2}\Big)[11-15]$
$=\Big(\frac{15}{2}\Big)(-4)$
$=(15)(-2)$
$=-30$
Therefore, the sum of the $15$ terms of the given A.P. is $S_{15} = -30.$
View full question & answer→Question 153 Marks
Find the number of natural numbers between $101$ and $999$ which are divisible by both $2$ and $5.$
AnswerSince, the number is divisible by both $2$ and $5$ , means it must be divisible by $10$ .
In the given numbers, first number that is divisible by $10$ is $110 $.
Next number is $110+10=120$.
The last number that is divisible by $10$ is $990 .$
Thus, the progression will be $110,120, \ldots . . ., 990$.
All the terms are divisible by 10 , and thus forms an A.P. having first term as $110$ and the common difference as $10$ .
We know that, $n^{\text {th }}$ term $=a_n=a+(n-1) d$
According to the question,
$990=110+(n-1) 10$
$\Rightarrow 990=110+10 n-10$
$\Rightarrow 10 n=990-100$
$\Rightarrow 10 n=890$
$\Rightarrow n=89$
Thus, the number of natural numbers $101$ and $999$ which are divisible by both $2$ and $5$ is $89 $.
View full question & answer→Question 163 Marks
Find the sum of the following arithmetic progressions:
$(x - y)^2, (x^2 + y^2), (x + y)^2, ......,$ to n terms.
AnswerIn an A.P. let first term = a, common difference = d, and there are n terms. Then, sum of n terms is,
$\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
A.P. is $(x - y)^2, (x^2 + y^2), (x + y)^2, .....,$ to n terms
Here,
$a = (x - y)^2, d =x^2 + y^2 - (x - y)^2 = x^2 + y^2 - x^2 - y^2 + 2xy$
$\Rightarrow d = 2xy$
$\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
$=\frac{\text{n}}{2}[2(\text{x}-\text{y})^2+(\text{n}-1)(-2\text{xy})]$
$=\frac{\text{n}}{2}[2(\text{x}-\text{y})^2-2(\text{n}-1)\text{xy}]$
$=\frac{\text{n}}{2}\times2[(\text{x}-\text{y})^2-(\text{x}-1)\text{xy}]$
$=\text{n}[(\text{x}-\text{y})^2-(\text{x}-1)\text{xy}]$
View full question & answer→Question 173 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{\text{n}-3}{3}$
Answer$\text{a}_\text{n}=\frac{\text{n}-3}{3}$
Here, the $n^{th}$ term is given by the above expression. So, to find the first term we use, $n = 1$, we get,
$\text{a}_1=\frac{(1)-2}{3}$
$=\frac{-1}{3}$
Similarly, we find the other four terms,
Second term (n = 2),
$\text{a}_1=\frac{(2)-2}{3}$
$=\frac{0}{3}$
$=0$
Third term (n = 3),
$\text{a}_3=\frac{(3)-2}{3}$
$=\frac{1}{3}$
Fourh terms (n = 4),
$\text{a}_4=\frac{(4)-2}{3}$
$=\frac{2}{3}$
Fifth term (n = 5),
$\text{a}_5=\frac{(5)-2}{3}$
$=\frac{3}{3}$
$=1$
Therefore, the first five terms for the given sequence are $\text{a}_1=\frac{-1}{3},\ \text{a}_2=0,\ \text{a}_3=\frac{1}{3},\ \text{a}_4=\frac{2}{3},\ \text{a}_5=1.$
View full question & answer→Question 183 Marks
Which term of the A.P.$ -2, -7, -12, …$ will be $-77$? Find the sum of this A.P. up to the term $-77$.
AnswerGiven,
$\text { A.P. }-2,-7,-12, \ldots . .$
Let the $n ^{\text {th }}$ term of an A.P. is $-77$ .
Then, first term $(a) = -2$
$\text { Common difference }(d)=-7-(-2)=-7+2=-5$
$\because n^{\text {th }} \text { term of an A.P., } T_n=a+(n-1) d$
$\Rightarrow-77=-2+(n-1)(-5)$
$\Rightarrow-75=-(n-1) \times 5$
$\Rightarrow(n-1)=15 \Rightarrow n=16$
So, the $16^{\text {th }}$ term of the given A.P. will be $-77$
Now, the sum of $n$ terms of an A.P. is
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
So, sum of 16 terms i.e. upto the term $-77$
$\text { i.e., } S_{16}=\frac{16}{2}[2 \times(-2)+( n -1)(-5)]$
$=8[-4+(16-1)(-5)]=8(-4-75)$
$=8 \times-79=-632$
Hence, the sum of this A.P. upto the term $-77$ is $-632$ .
View full question & answer→Question 193 Marks
If the sum of first $p$ term of an A.P. is $a p^2+b p$, find its common difference.
AnswerGiven,
Sum of $p$ terms, $S_p=a p^2+b p$
Putting $p=1,2,3,4$,....
$S_1=a(1)^2+b(1)=a \times 1+b=a+b$
$S_2=a(2)^2+b(2)=a \times 4+2 b=4 a+2 b$
$S_3=a(3)^3+b(3)=a \times 9+3 b=9 a+3 b$
And $S_4=a(4)^2+b(4)=a \times 16+4 b=16 a+4 b$
We know $a_n=S_n-S_{n-1}$
$2^{\text {nd }}$ term, $a _2= S _2- S _1$
$\Rightarrow a _2=4 a +2 b-( a + b )$
$\Rightarrow a_2=4 a+2 b-a-b$
$\Rightarrow a_2=3 a+b$
$3^{\text {rd }}$ term, $a_3=S_3-S_2$
$\Rightarrow a _3=9 a +3 b-(4 a -2 b)$
$\Rightarrow a_3=9 a+3 b-(4 a+2 b)$
$\Rightarrow a_3=5 a+b$
$4^{\text {th }}$ term, $a_4= S _4- S _3$
$\Rightarrow a_4=16 a+4 b-(9 a+3 b)$
$\Rightarrow a_4=16 a+4 b-9 a-3 b$
$\Rightarrow a_4=7 a+b$
Now difference between two terms,
$d_1=a_3-a_2=5 a+b-(3 a+b)$
$=5 a+b-3 a-b$
$\Rightarrow d_1=2 a$
$d_2=a_4-a_3=7 a+b-(5 a+b)$
$=7 a+b-5 a-b$
$\Rightarrow d_2=2 a$
Hence, common difference is $2$ a .
View full question & answer→Question 203 Marks
How many numbers lie between $10$ and $300$, which when divided by $4$ leave a remainder 3?
AnswerHere, the first number is $11$, which divided by $4$ leave remainder $3$ between $10$ and $300$.
Last term before $300$ is $299$, which divided by 4 leave remainder 3.
$11, 15, 19, 23, ...., 299.$
Here, first term $(a) = 11,$
Common differnce $(d) = 15 - 11 = 4$
$n^{th}$ term, $a_n = a + (n - 1)d = l$[last term]
$\Rightarrow 299 = 11 + (n - 1)4$
$\Rightarrow 299 - 11 = (n - 1)4$
$\Rightarrow 4(n - 1) = 288$
$\Rightarrow (n - 1) = 72$
$n = 73.$
View full question & answer→Question 213 Marks
Find:$n^{th}$ term of the A.P. $13, 8, 3, -2, ....$
AnswerGiven A.P., $13, 8, 3, -2, .....$
Here,
First term, $a = 13$
Difference, $d = (8 - 13) = -5$
We have to find $n^{th}$ term,
So putting $n = n$
We know, $n^{th}$^ term of A.P.
$a_n = a + (n - 1)d$
$\Rightarrow a_n = 13 + (n - 1)(-5)$
$\Rightarrow a_n = 13 + (-5n + 5)$
$\Rightarrow a_n = 13 - 5n + 5$
$\Rightarrow a_n = 18 - 5n$
Hence, $n^{th}$ term of given A.P. is $18 - 5n.$
View full question & answer→Question 223 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$1^2, 3^2, 5^2, 7^2, .....$
AnswerIn the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference (d),
Given,
$1^2, 3^2, 5^2, 7^2, .....$
Here, $a_1 = 1^2, a_2 = 3^2, a_3 = 5^2, a_4 = 7^2$
Difference between terms,
$d_1 = a_2 - a_1 = 3^2 - 1^2 = 9 - 1 = 8,$
$d_2 = a_3 - a_2 = 5^2 - 3^2 = 25 - 9 = 16,$
and $d_3 = a_4 - a_3 = 7^2 - 5^2 = 49 - 25 = 24$
Hence, difference $d_1, d_2$_ and $d_3$_ are not equal, trem sequence not in an A.P.
View full question & answer→Question 233 Marks
Show that the sequence defined by $a_n = 3n^2 - 5$ is not an A.P.
AnswerGiven sequence is,
$a_n = 3n^2 - 5.$
$n^{th}$ term of given sequence $(a_n) = 3n^2 - 5.$
$(n + 1)^{th} $term of given sequence $(a_n + 1) = 3(n + 1)^2 - 5$
$= 3(n^2 + 1^2 + 2n.1) - 5$
$= 3n^2 + 6n - 2$
$\therefore$ The common difference $(d) = a_n + 1 - an$
$d = (3n^2 + 6n - 2) - (3n^2 - 5)$
$= 3a^2 + 6n - 2 - 3n^2 + 5$
$= 6n + 3$
Common difference $(d)$ depends on $'n'$ value
$\therefore$ Given sequence is not in A.P.
View full question & answer→Question 243 Marks
The $9^{\text {th }}$ term of an A.P. is equal to $6$ times its second term. If its $5^{\text {th }}$ term is $22$ , find the A.P.?
AnswerLet $a$ be the first term and be the common difference and
$T_n=a+(n-1) d$
$\therefore T_9=a+(9-1) d=a+8 d$
$T_2=a+(2-1) d=a+d$
According to question
$T_9=6 T_2$
$a+8 d=6(a+d)$
$a+8 d=6 a+6 d$
$\Rightarrow 8 d-6 d=6 a-a \Rightarrow 5 a=2 d$
$\Rightarrow\ \text{a}=\frac{2}{5}\text{d}\ .....(\text{i})$
and $\text{T}_5=\text{a}+(5-1)\text{d}=\text{a}+4\text{d}$
$\therefore\ \text{a}+4\text{d}=22$
$\Rightarrow\ \frac{2}{5}\text{d}+4\text{d}=22\ [\text{From (i)}]$
$\Rightarrow\ 2\text{d}+20\text{d}=22\times5\Rightarrow\ 22\text{d}=22\times5$
$\Rightarrow\ \text{d}=\frac{22\times5}{22}=5$
$\therefore\ \text{a}=\frac{2}{5}\text{d}=\frac{2}{5}\times5=2$
$\therefore\ \text{A.P.}=2,7,12,17, .....$
View full question & answer→Question 253 Marks
Find the common difference and write the next four terms of the following arithmetic progressions:
$1, -2, -5, -8, .....$
AnswerHere, $a_1 = 1, a_2 = -2, a_3 = -5, a_4 = -8, .....$
Now $a_2 - a_1 = -2 -1 = -3$
$a_3 - a_2 = -5 - (-2) = -5 + 2 = -3$
$a_4 - a_3 = -8 - (-5) = -8 + 5 = -3$
$\therefore$ It is an A.P. whose common difference is $= -3$
Now next four terms will be
$-8 - 3 = -11$
$-11 - 3 = -14$
$-14 - 3 = -17$
$-17 - 3 = -20$
$\therefore$ $-11, -14, -17, -20$ are four term next to these.
View full question & answer→Question 263 Marks
The sum of the first n terms of an A.P. is $3n^2 + 6n$.
Find the $n^{th}$ term of this A.P.
AnswerSum of n terms $(S_n) = 3n^2 + 6n$
Let $T_n$_ of $a_n$_ be the $n^{th}$^ term, then
$a_n = S_n - S_{n-1}$
$= (3n^2 + 6n) - {3(n - 1)^2 + 6(n - 1)}$
$= (3n^2 + 6n) - {3(n^2 - 2n + 1 + 6n - 6)}$
$= (3n^2 + 6n) - (3n^2 - 6n + 3 + 6n - 6)$
$= 3n^2 + 6n - 3n^2 + 6n - 3 - 6n + 6$
$= 6n + 3$
View full question & answer→Question 273 Marks
Justify whether it is true to say that the sequence, having following $n^{th}$ term is an A.P.
$a_n = 2n - 1.$
AnswerYes, here $a_n = 2n - 1$
Put $n = 1, a_1 = 2(1) - 1 = 1$
Put $n = 2, a_2 = 2(2) - 1 = 3$
Put $n = 3, a_3 = 2(3) - 1 = 5$
Put $n = 4, a_4 = 2(8) - 1 = 7$
List of numbers becomes $1, 3, 5, 7, .....$
Here,
$a_2 - a_1 = 3 - 1 = 2$
$a_3 - a_2 = 5 - 3 = 2$
$a_4 - a_3 = 7 - 5 = 2$
$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = .....$
Hence, $2n - 1$ is the $n^{th}$ term of A.P.
View full question & answer→Question 283 Marks
An A.P. consists of 60 terms. If the first and the last terms be $7$ and $125$ respectively, find the $32^{\text {nd }}$ term.
AnswerGiven
No of terms $= n =60$
First term (a) = 7
Last term $a_{60}=125$
$a_{60}=a+(60-1) \times d\left(\therefore a_n=a+(n-1) d\right)$
$125=7+59 \times d$
$118=59 d$
$d =\frac{118}{59}=2$
$32^{\text {nd }}$ term $a_{32}=a+(32-1) d$
$=7+31 \times 2$
$=7+62$
$=69$
View full question & answer→Question 293 Marks
The sum of first $n$ terms of an A.P. is $5 n-n^2$. Find the $n^{\text {th }}$ term of this A.P.
AnswerLet $a$ be the first term and $d$ be the common difference.
We know that, sum of first n terms $= S _{ n }=\frac{ n }{2}[2 a +( n -1) d ]$
It is given that sum of the first $n$ terms of an A.P. is $5 n-n^2$.
$\therefore$ First term $= a = S _1=5(1)-(1)^2=4$.
Sum of first two terms $=S_2=5(2)-(2)_2=6$.
$\therefore$ Second term $= S _2- S _1=6-4=2$.
$\therefore$ Common difference $= d =$ Second term - First term
$=2-4=-2$
Also, $n^{\text {th }}$ term $=a_n=a+(n-1) d$
$\Rightarrow a_n=a+(n-1)(-2)$
$\Rightarrow a_n=4-2 n+2$
$\Rightarrow a_n=6-2 n$
Thus, $n ^{\text {th }}$ term of this A.P. is $6-2 n$.
View full question & answer→Question 303 Marks
Solve the question: $(-4) + (-1) + 2 + 5 + ..... + x = 437.$
AnswerSuppose $x$ is $n^{th}$ term of the given A.P.
$a_n = x$
Here, $a = -4, d = 3.$
It is given that, $S_n = 437.$
$\Rightarrow\ \frac{\text{n}}{2}[2(-4)+(\text{n}-1)3]=437$
$\Rightarrow\ 3\text{n}^2-11\text{n}-874=0$
$\Rightarrow\ 3\text{n}^2-57\text{n}+46\text{n}-874=0$
$\Rightarrow\ 3\text{n}(\text{n}-19)+46(\text{n}-19)=0$
$\Rightarrow\ \text{n}=-\frac{46}{3}, 19$
Since, n cannot be in fraction so $n = 19.$
Now $a_n = x$
$\Rightarrow (-4) + (19 - 1)3 = x$
$\Rightarrow -4 + 54 = x$
$\Rightarrow x = 50.$
View full question & answer→Question 313 Marks
Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.
AnswerLet the four terms of the A.P. be a - 3d, a - d, a + d and a + 3d.
Given:
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 56
⇒ 4a = 56
⇒ a = 14
Also,
$\frac{(\text{a}-3\text{d})(\text{a}+3\text{d})}{(\text{a}-\text{d})(\text{a}+\text{d})}=\frac{5}{6}$
$\Rightarrow\ \frac{\text{a}^2-9\text{d}^2}{\text{a}^2-\text{d}^2}=\frac{5}{6}$
$\Rightarrow\ \frac{(14)^2-9\text{d}^2}{(14)^2-\text{d}^2}=\frac{5}{6}$
$\Rightarrow\ \frac{196-9\text{d}^2}{196-\text{d}^2}=\frac{5}{6}$
$\Rightarrow\ 1176 - 54\text{d}^2=980-5\text{d}^2$
$\Rightarrow\ 196=49\text{d}^2$
$\Rightarrow\ \text{d}^2=4$
$\Rightarrow\ \text{d}=\pm2$
When d = 2, the terms of the AP are 8, 12, 16, 20. When d = -2, the terms of the AP are 20, 18, 12, 8.
View full question & answer→Question 323 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$1^2, 5^2, 7^2, 73, .....$
AnswerIn the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference (d), Here,
First term (a) $=1^2$
$a_1=5^2$
$a_2=7^2$
Now, for the given to sequence to be an A.P.
Common difference $(d)=a_1-a=a_2-a_1$
Here,
$a_2-a_1=5^2-1^2$
$=25-1$
$=24$
Also,
$a_3-a_2=7^2-5^2=49-25=24$
Since $a_1-a=a_2-a_1$
Hence, the given sequence is an A.P. with the common difference $d=24$.
View full question & answer→Question 333 Marks
How many numbers of two digit are divisible by $3$?
AnswerTwo digit numbers divisible by 3
are, $12, 15, 18, ....., 99$
Hence, First term $a = 12$
Difference $d = 15 - 12 = 3$
and Last term a_n = 99
We know $n^{th}$^ term of an A.P.
$a_n = a + (n - 1)d$
$\Rightarrow 99 = 12 + (n - 1)3$
$\Rightarrow 99 = 12 + 3n - 3$
$\Rightarrow 99 = 9 + 3n$
$\Rightarrow 90 = 3n$
$\Rightarrow n = 30$
Hence, Total number of two digit which one divisible by $3$ is $30.$
View full question & answer→Question 343 Marks
All integers from $1$ to $500$ which are multiplies $2$ as well as of $5$.
AnswerSince, multiples of $2$ as well as of $5=$ LCM
$\text { of }(2,5)=10$
Multiples of 2 as wekk as of $5$ from $1$ to $500$ is $10,20,30, \ldots . . ., 500$
$\therefore a=10, d=10, a_n=l=500$
$\because a_n=a+(n-1) d=1$
$\Rightarrow 500=10+(n-1) 10$
$\Rightarrow 490=(n-1) 10$
$\Rightarrow n-1=49 \Rightarrow n=50$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\Rightarrow\ \text{S}_{50}=\frac{50}{2}(10 + 500)=\frac{50}{2}\times510$
$=50\times255=12750$
View full question & answer→Question 353 Marks
If $9^{\text {th }}$ term of an A.P is zero, prove that its $29^{\text {th }}$ term is double the $19^{\text {th }}$ term.
AnswerGiven
$9^{\text {th }}$ term of $a n$ A.P. $a g=0 a_n=a+(n-1) d$
$a+(9-1) d=0$
$a+8 d=0$
$a=-8 d$
We have to prove
$29^{\text {th }}$ term is double the $19^{\text {th }}$ term $a_{29}=2 \cdot a_{19}$
$a+28 d=2[a+18 a]$
Put $a=-8 d$
$-8 d+28 d=2[-8 d+18 d]$
$20 d=2 \times 10 d$
$20 d=20 d$
Hence proved.
View full question & answer→Question 363 Marks
In an A.P., the first term is $2$, the last term is $29$ and the sum of the terms is $155$. Find the common difference of the A.P.
AnswerIn the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.
Here,
The first term of the A.P $(a) = 2$
$$The last term of the A.P $(l) = 29$
Sum of all the terms $(S_n) = 155$
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
$155=\Big(\frac{\text{n}}{2}\Big)(2+29)$
$155=\Big(\frac{\text{n}}{2}\Big)(31)$
$155(2)=\text{(n)}(31)$
$\text{n}=\frac{310}{31}$
$\text{n}=10$
Now, to find the common difference of the A.P. we use the following formula,
$l = a + (n - 1)d$
We get
$29 = 2 + (10 - 1)d$
$29 = 2 + (9)d$
$29 - 2 = 9d$
$\text{d}=\frac{27}{9}$
$d = 3$
Therefore, the common difference of the A.P. is $d = 3.$
View full question & answer→Question 373 Marks
Find the indicated terms in the following sequences whose $n^{\text {th }}$ terms are:
$a_n=(n-1)(2-n)(3+n) ; a_1, a_2, a_3$.
Answer$a_n=(n-1)(2-n)(3+n)$
We need to find $a_1, a_2$ and $a_3$
Now, to find $a_1$ term we use $n =1$, we get,
$a_1=(1-1)(2-1)(3+1)$
$=(0)(1)(4)$
$=0$
Also, to find $a_2$ term we use $n=2$, we get
$a_2=(2-1)(2-2)(3-2)$
$=(1)(0)(5)$
$=0$
Similarly, to find as term we use $n=3$, we get,
$a_3=(3-1)(2-3)(3+3)$
$=(2)(-1)(6)$
$=-12$
Thus, $a_1=0, a_2=0$ and $a_3=-12$.
View full question & answer→Question 383 Marks
How many terms are there in the A.P.?
$7, 13, 19, ....., 205.$
AnswerGiven,A.P. $7, 13, 19, ....., 205$
Here,
First term, $a =7$,
Difference $d=13-7=6$
Last $n ^{\text {th }}$ term $a _{ n }=205$
We know, $n ^{\text {th }}$ term of A.P.
$a_n=a+(n-1) d$
$\Rightarrow 205=7+(n-1) 6$
$\Rightarrow 205=7+6 n-6$
$\Rightarrow 205=1+6 n$
$\Rightarrow 6 n=204$
$\Rightarrow n=\frac{204}{6}$
$\Rightarrow n=34$
Hence, Total $34$ terms in given A.P.
View full question & answer→Question 393 Marks
For the following arithmetic progressions write the first term a and the common difference d:
$-5, -1, 3, 7, ....$
AnswerA.P. is, $-5, -1, 3, 7, .......$
Here,
First term $a = -5$
Common difference,
$a_1 - a = -1 - (-5) = 4$
$a_2 - a_1 = 3 - (-1) = 4$
$d = 4$
Therefore $a = -5$ and $d = 4.$
View full question & answer→Question 403 Marks
Write the sum of first n odd natural numbers.
AnswerLet,
Odd numbers are $1, 3, 5, 7, ....., n$
Here,
First term $a = 1$
Difference $d = 3 - 1 = 2$
We know, Sum of n terms
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(1)+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2+2\text{n}-2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2\text{n}$
$\Rightarrow\ \text{S}_\text{n}=\text{n}^2$
Hence, Sum of first n odd numbers is $n^2$.
View full question & answer→Question 413 Marks
Justify whether it is true to say that the sequence, having following $n^{th}$ term is an A.P.
$a_n = 3n^2 + 5.$
AnswerNo, here $a_n = 3n^2 + 5$
Put $n = 1, a_1 = 3(1)^2 + 5 = 8$
Put $n = 2, a_2 = 3(2)^2 + 5 = 3(4) + 5 = 17$
Put $n = 3, a_3 = 3(3)^2 + 5 = 3(9) + 5 = 27 + 5 = 32$
So, the list of number becomes $8, 17, 32, .....$
Here,
$a_2 - a_1 = 17 - 8 = 9$
$a_3 - a_2 = 32 - 17 = 15$
$\therefore\ \text{a}_2-\text{a}_1\neq\text{a}_3-\text{a}_2$
Since, the successive difference of the list is not same. So, it does not from an A.P.
View full question & answer→Question 423 Marks
How many three digit numbers are divisible by 7?
AnswerThe three digit numbers are 100, ..... 999, 105 is the first 3 digit number which is divisible by 7 when we divide 999 with 7 remainder is 5. So, 999 - 5 = 994 is the last three digits divisible by 7 so, the sequence is
105, ....., 994
First term (a) = 105
Last term (1) = 994
Common difference (d) = 7
Let there are n numbers in the sequence
an = 994
a + (n - 1)d = 994
a + (n - 1)d = 994
105 + (n - 1)7 = 994
(n - 1)7 = 889
$\text{n}-1=\frac{889}{7}=127$
n = 128
$\therefore$ There are 128 numbers between 105, 994 which are divisible by 7.
View full question & answer→Question 433 Marks
Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find.
$S _{22}$, if $d =22$ and $a _{22}=149$
AnswerGiven $d =22, a _{22}=149, n =22$
We know that
$a_n=a+(n-1) d$
$149=a+(22-1) 22$
$149=a+462$
$a=-313$
Now, Sum is given by
$S_{n}=\frac{n}{2}[2 a(n-1) d]$
Where; a = first term for the given A.P.
$d = common$ difference of the given A.P.
$n =$ number of terms
So, using the formula for $n =22$, we get
$\left.S_{22}=\frac{22}{2}\{2 \times(-313)+(22-1) \times 22)\right\}$
$S_{22}=11\{-626+462\}$
$S_{22}=-1804$
Hence, the sum of $22$ terms is $-1804$ .
View full question & answer→Question 443 Marks
Find the next five terms of the following sequences given:
$\text{a}_1=1,\text{a}_\text{n}=\text{a}_{\text{n}-1}+2,\text{n}\geq2.$
Answer$a_1 = 1, a_n - a_{n-1} + 2$
Let $n = 2, 3, 4, 5, 6$
$\therefore$ $a_2 = a_{2-1} + 2 = a_1 + 2$
$= 1 + 2 = 3$ ($\because$ $a_1 = 1$)
$a_3 = a_{3-1} + 2 = a_2 + 2$
$= 3 + 2 = 5$
$a_4 = a_{4-1} + 2 = a_3 + 2$
$= 5 + 2 = 7$
$a_5 = a_{5-1} + 2 = a_4 + 2$
$= 7 + 2 = 9$
$a_6 = a_{6-1} + 2 = a_5 + 2$
$= 9 + 2 = 11.$
View full question & answer→Question 453 Marks
Find the common difference and write the next four terms of the following arithmetic progressions:
$0,-3,-6,-9, \ldots . .$
AnswerHere, first term $\left(a_1\right)=0$
$\text { Common difference }(d)=a_2-a_1$
$=-3-0$
$=-3$
Now, we need to find the next four terms of the given A.P.
That is we need to find $a_5, a_6, a_7, a_8$
So, using the formula $a_n=a+(n-1) d$
Substituting $n=5,6,7,8$ in the above formula
Substituting $n=5$, we get
$a_5=0+(5-1)(-3)$
$a_5=0-12$
$a_5=-12$
Substituting $n =6$, we get
$a_6=0+(6-1)(-3)$
$a_6=0-15$
$a_6=-15$
Substituting $n =7$, we get
$a_7=0+(7-1)(-3)$
$a_7=0-18$
$a_7=-18$
Substituting $n =8$, we get
$a_8=0+(8-1)(-3)$
$a_8=0-21$
$a_8=-21$
Therefore, the common difference is $d=-3$ and the next four terms are $-12,-15,-18,-21$.
View full question & answer→Question 463 Marks
Find the sum of first $51$ terms of an A.P. whose second and third terms are $14$ and $18 $respectively.
AnswerGiven, $a_2 = 14 \Rightarrow a + d = 14 .....(i)$
$a_3 = 18 \Rightarrow a + 2d = 18 .....(ii)$
Put $d = 4$ is (i) $a + 4 = 14 a = 10$
$\therefore\ \text{S}_{51}=\frac{51}{2}\{2\times10+(51-1)\times4\}$
$\Big(\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}\Big)$
$=\frac{51}{2}\{20+200\}$
$=\frac{51}{2}\times220$
$=5610$ $\therefore\ \text{S}_{51}=5610$ View full question & answer→Question 473 Marks
The sum of the first n terms of an A.P. is $ 4n^2 + 2n.$ Find the $n^{th}$ term of this A.P.
Answer$S_n = 4n^2 + 2n$
$\therefore$ $a_n = S_n - S_{n-1}$
$= (4n^2 + 2n) - [4(n - 1)^2 + 2(n - 1)]$
$= (4n^2 + 2n) - [4(n^2 - 2n + 1) + 2n - 2]$
$= (4n^2 + 2n) - [4n^2 - 8n + 4 + 2n - 2]$
$= 4n^2 + 2n - 4n^2 + 8n - 4 - 2n + 2$
$= 8n - 2 = 2(4n - 1)$
View full question & answer→Question 483 Marks
Ramkali would need Rs. 1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved Rs. 50 in the first month of this year and increased her monthly saving by Rs. 20. After a year, how much money will she save? Will she be able to fulfil her dream of sending her daughter to school?
AnswerAdmission fee and books etc. = Rs. 1800
First month's savings = Rs. 50
Increase in monthly savings = Rs. 20
Period = 1 year = 12 months
Here a = 50, d = 20 and n = 12
$\text{S}_{12}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{12}{2}[2\times50+(12-1)\times20]$
= 6[100 + 11 × 20]
= 6[100 + 220]
= 6 × 320 = Rs. 1920
Savings = Rs. 1920
Yes, she will be able to send her daughter.
View full question & answer→Question 493 Marks
If $\frac{4}{5}$, $a, 2$ are three consecutive terms of an A.P.?
AnswerHere, we are given three consecutive terms of an A.P.
First term $(a_1)$ = $\frac{4}{5}$
Second term $(a_2) = a$
Third term $(a_3) = 2$
We need to find the value of a. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
$d = a_2 - a_1$
$\text{d}=\text{a}-\frac{4}{5}\ ....\text{(i)}$
Also,
$d = a_3 - a_2$
$d = 2 - a$
Now, on equating (i) and (ii), we get,
$\text{a}-\frac{4}{5}=2-\text{a}$
$\text{a}+\text{a}=2+\frac{4}{5}$
$2\text{a}=\frac{10+4}{5}$
$\text{a}=\frac{14}{10}$
$\text{a}=\frac{7}{5}$
Therefore, $\text{a}=\frac{7}{5}$.
View full question & answer→Question 503 Marks
Find the common difference of the A.P. and write the next two terms:
$51, 59, 67, 75, .....$
Answer$51,59,67,75, \ldots .$.
Here,
$a_1=51$
$a_2=59$
So, common difference of the A.P. (d) $=a_2-a_1$
$=59-51$
$=8$
Also, we need to find the next two terms of A.P., which means we have to find the $5^{\text {th }}$ and $6^{\text {th }}$ term. So, for fifth term,
$a_5=a_1+4 d$
$=51+4(8)$
$=51+32$
$=83$
Similarly, we find the sixth term,
$a_6=a_1+5 d$
$=51+5(8)$
$=51+40$
$=91$
Therefore, the common difference is $d=8$ and the next two terms of the A.P. are $a_5=83, a_6=91$.
View full question & answer→Question 513 Marks
Two A.P.s have the same common difference. The first term of one A.P. is $2$ and that of the other is $7$ . The difference between their $10^{\text {th }}$ terms is the same as the difference between their $21^{\text {st }}$ terms, which is the same as the difference between any two corresponding terms. Why?
AnswerFirst term of $1^{\text {st }}$ A.P. is 2 .
First term of $2^{\text {nd }} A . P$. is 7.
Consider the difference of their $10^{\text {th }}$ terms.
$a_{10}-a^{\prime}{ }_{10}=a+9 d-a^{\prime}-9 d^{\prime}$
$=a-a^{\prime}+9 d-9 d^{\prime}$
$=2-7+0\left[d=d^{\prime}\right]$
$=-5$
$a_{21}-a^{\prime} 21=a+20 d-a^{\prime}-20 d^{\prime}$
$=a-a^{\prime}+20 d-20 d^{\prime}$
$=2-7+0\left[d=d^{\prime}\right]$
$=-5$
Therefore, $a_{10}-a^{\prime}{ }_{10}=a_{21}-a^{\prime}{ }_{21}$
The difference between any two corresponding terms of A.P's in same as the difference between their terms.
View full question & answer→Question 523 Marks
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are $1$ to $12$ classes in the school and each class has two sections, find how many trees were planted by the students.
AnswerClasses: $I + II + III + .... + XII$
Sections: $2(I) + 2(II) + 2(III) + ..... + 2(XII)$
Total nos. of trees $= (2 \times 2) + (2 \times 4) + (2 \times 6) + ..... + (2 \times 24) = 4 + 8 + 12 + ..... + 48$
Here $a_1 = 4$ $\therefore\ \text{S}_{12}=\frac{12}{2}(4+48)$ $= 6(52) n = 12 = 312\ trees$
Here $a_1 = 4 a_n = 48 n = 12$ $\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}_1+\text{a}_\text{n})$
Value: We should care for our environment and love nature.
View full question & answer→Question 533 Marks
Write the value of $a_{30} - a_{10}$_ for the A.P. $4, 9, 14, 19, .....$
AnswerGiven,
A.P. $4, 9, 14, 19, .....$
Here, First term $a = 4$
and Difference $d = 9 - 4 = 5$
We know,
$a_n = a + (n - 1)d$
$30^{th}$ term
$a_{30} = 4 + (30 - 1)5$
$\Rightarrow a_{30} = 4 + 29 \times 5$
$\Rightarrow a_{30} = 4 + 145$
$\Rightarrow a_{30} = 149$
$10^{th}$^ term,
$a_{10} = 4 + (10 - 1)5$
$\Rightarrow a_{10} = 4 + 9 \times 5$
$\Rightarrow a_{10} = 4 + 45$
$\Rightarrow a_{10} = 49$
Now, we have to find $a_{30} - a_{10}$_
Now, we have to find $a_{30} - a_{10}$_
$\Rightarrow a_{30} - a_{10} = 149 - 49$
$\Rightarrow a_{30} - a_{10} = 100$
Hence, value of $a_{30} - a_{10}$_ is $100.$
View full question & answer→Question 543 Marks
Write the expression $a_n - a_k$_ for the A.P. $a, a + d, a + 2d, ...$
Hence, find the common difference of the A.P. for which,
$a_{10} - a_5 = 200.$
AnswerWe know,
$a_n = a + (n - 1)d$
Let,
$n^{th}\text {term}, a_n = a + (n - 1)d$
$\Rightarrow a_n = a + nd - d$
$k^{th}\text {term}, a_k = a + (k - 1)d$
$\Rightarrow a_k = a + kd - d$
Now,
$\Rightarrow a_n - a_k = (a + nd - d) - (a + kd - d)$
$\Rightarrow = a + nd - d - a - kd + d$
$\Rightarrow = nd - kd$
$\Rightarrow = d(n - k)$
Given,
$a_{10} - a_5 = 200$
From (1) $a_{10} - a_5 = (10 - 5)d$
$200 = 5 × d$
$\text{d}=\frac{200}{5}=40\Rightarrow\ \text{d}=40$
View full question & answer→Question 553 Marks
How many terms of the sequence $18, 16, 14, .....$ should be taken so that sum is zero?
AnswerGiven A.P.
$18, 16, 14, .....$
Sum, $S_n = 0$
Here, First term $a = 18$
and Difference $d = 16 - 18 = -2$
We know,
$\text{S}_\text{n}=\frac{\text{n}}{2}\Big[2\text{a}+(\text{n}-1)\text{d}\Big]$
$\Rightarrow\ 0=\frac{\text{n}}{2}[2(18)+(\text{n}-1)(-2)]$
$\Rightarrow 0 = n[36 - 2n + 2]$
$\Rightarrow 0 = n[38 - 2n]$
$\Rightarrow 0 = 38 - 2n$
$\Rightarrow 2n = 38$
$\Rightarrow n = 19$
Hence, Sum of $19$ terms is zero.
View full question & answer→Question 563 Marks
Let there be an A.P. with first term $'a'$, common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms,
find. a , if $a _{ n }=28, S_{ n }=144$ and $n =9$.
Answer$a_n=28, S_n=144, n=9$
$a_n=a+(n-1) d$
$\Rightarrow a_9=a+(9-1) d=a+8 d$
$a+8 d=28 \ldots . .(i)$
$S_n=\frac{n}{2}[a+1] \Rightarrow \frac{9}{2}(a+28)=144$
$\Rightarrow a+28=\frac{144 \times 2}{9}=32$
$a=32-48=4$
$\therefore \text { From (i) }$
$4+8 d=28 \Rightarrow 8 d=28-4=24$
$\Rightarrow a=\frac{24}{8}=3$
Hence $a=4$.
View full question & answer→Question 573 Marks
How many multiples of $4$ lie between $10$ and $250$?
AnswerLet,
Multiple of $4$ lie between $10$ and $250$
$12, 16, 20, ..... 248$
we know $a_n=a+(n-1) d$
Here,
First term $a =12$
Difference $d=16-12=4$
and Last $n ^{\text {th }}$ term $a _{ n }=248$
Then, $a_n=a+(n-1) d$
$\Rightarrow 248=12+(n-1) 4$
$\Rightarrow 248=12+4 n-4$
$\Rightarrow 4 n =248-12+4$
$\Rightarrow 4 n =240$
$\Rightarrow n =60$
Hence, multiple of $4$ lies between $10$ and $250$ is $60$ .
View full question & answer→Question 583 Marks
Write the common difference of an A.P. whose $n^{th}$ term is $a_n = 3n + 7.$
AnswerGiven,
$a_n = 3n + 7$
Puting $n = 1, 2, 3, .....$
$a_1 = 3(1) + 7 = 3 + 7 = 10,$
$a_2 = 3(2) + 7 = 6 + 7 = 13,$
and $a_3 = 3(3) + 7 = 9 + 7 = 16$
Now, Common difference
$d = a_2 - a_1 = 13 - 10 = 3$
$d = a_3 - a_2 = 16 - 13 = 3$
Hence, common difference is $3$.
View full question & answer→Question 593 Marks
Find the sum of all integers between $100$ and $550,$ which are divisible by $9$.
AnswerAll integers between $100$ and $550$ which are divisible by $9$ are
$108, 117, 126, 135, ....., 549$
Where $a = 108, d = 9$ and $l = 549$
$\therefore$$a_n = a + (n - 1)d$
$\Rightarrow 549 = 108 + (n - 1) \times 9$
$\Rightarrow 549 = 108 + 9n - 9$
$\Rightarrow 9n = 549 - 108 + 9 = 558 - 108 = 450$
$\therefore\ \text{n}=\frac{450}{9}=50$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\Rightarrow\ \text{S}_{20}=\frac{50}{2}[108+549]$
$=25(657)=16425$
View full question & answer→Question 603 Marks
Write $5^{th}$ term from the end of the A.P. 3$, 5, 7, 9, ....., 201$
AnswerGiven,
A.P, $3, 5, 7, 9, ..... 201$
Here, First term $a=3$
Difference $d =5-3=2$
and Last term $a _{ n }=201$
We knaw,
$a_n=a+(n-1) d$
$\Rightarrow 201=3+(n-1) 2$
$\Rightarrow 201=3+2 n-2$
$\Rightarrow 201=1+2 n$
$\Rightarrow 2 n=201-1$
$\Rightarrow 2 n=200$
$\Rightarrow n=\frac{100}{2}$
$\Rightarrow n=100$
Now, we have to find $5^{\text {th }}$ term from the end $100^{\text {th }}-4^{\text {th }}=96^{\text {th }}$
$a_n=a+(n-1) d$
$\Rightarrow a_{96}=3+(96-1) 2$
$\Rightarrow a_{96}=3+95 \times 2$
$\Rightarrow a_{96}=3+190$
$\Rightarrow a_{96}=193$
Hence, $5^{\text {th }}$ term from the end of given A.P. is $193.$
View full question & answer→Question 613 Marks
The first and the last terms of an A.P. are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum?
AnswerGiven,
$a = 17, l = 350, d = 9$
$l = a_n = a + (n - 1)d$
$350 = 17 + (n - 1)9$
$333 = (n - 1)9$
$\text{n}-1=\frac{333}{9}=37$
$\text{n}=38$
$\therefore$ 38 terms ate there
$\text{S}_\text{n}=\frac{\text{n}}{2}\{\text{a}+\text{l}\}$
$\text{S}_{38}=\frac{38}{2}(17+350)$
$=19\times367$
$\therefore\ \text{S}_\text{38}=6973$
View full question & answer→Question 623 Marks
In which of the following situations, the sequence of numbers formed will form an A.P.?
Divya deposited $Rs. 1000$ at compound interest at the rate of $10\%$ per annum. The amount at the end of first year, second year, third year, ...., and so on.
AnswerAmount ar the end of the $1^{st}$ year = $Rs. 1100$
Amount at the end of the $2^{nd}$ year = $Rs. 1210$
Amount at the end of $3^{rd}$ year = $Rs. 1331$ and so on.
So, the amount (in Rs.) at the end of $1^{st}$ year, $2^{nd}$ year, $3^{rd}$ year, ..... are
$1100, 1210, 1331, .....$
Here, $a_2 - a_1 = 110$
$a_3 - a_2 = 121$
As $\text{a}_2-\text{a}_1\neq\text{a}_3-\text{a}_2,$ it does not from an AP.
View full question & answer→Question 633 Marks
For the following arithmetic progressions write the first term a and the common difference d:
$0.3, 0.55, 0.80, 1.05, ....$
Answer$0.3, 0.55, 0.80, 1.05, .....$
Here first term $(a) = 0.3$
We have $= a_2 - a_1 = 0.55 - 0.3 = 0.25$
$a_3 - a_2 = 0.80 - 0.55 = 0.25$
$\therefore$ Common difference $= 0.25.$
View full question & answer→Question 643 Marks
Sum of the first 14 terms of an A.P. is $1505$ and its first term is $10.$ Find its $25^{th}$^ term.
AnswerSum of first $14$ terms of an A.P. $= 1505$
First term $(n) = 10$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{14}=\frac{14}{2}[2\times10+(14-1)\text{d]}$
$\Rightarrow\ 1505=7(20+13\text{d})$
$\Rightarrow\ 1505=140+91\text{d}$
$\Rightarrow\ 91\text{d}=1505-140=1365$
$\text{d}=\frac{+1365}{91}=+15$
$\therefore\ \text{T}_{25}=\text{a}+(\text{n}-1)\text{d}$
$=10+(25-1)(15)$
$=10+24\times(+15)=10+360=370$
View full question & answer→Question 653 Marks
The sum of $5^{\text {th }}$ and $9^{\text {th }}$ terms of an A.P. is $30$ . If its $25^{\text {th }}$ term is three times its $8^{\text {th }}$ term, find the A.P.?
AnswerLet a be the first term and d be the common difference.
We know that, $n^{\text {th }}$ term $=a_n=a+(n-1) d$
According to the question,
$a_5+a g=30$
$\Rightarrow a+(5-1) d+a+(9-1) d=30$
$\Rightarrow a+4 d+a+8 d=30$
$\Rightarrow 2 a+12 d=30$
$\Rightarrow a+6 d=15 \ldots . .(1)$
Also,
$a_{25}=3\left(a_8\right)$
$\Rightarrow a+(25-1) d=3[a+(8-1) d]$
$\Rightarrow a+24 d=3 a-21 d$
$\Rightarrow 3 a-a=24 d-21 d$
$\Rightarrow 2 a=3 d$
$a=\frac{3}{2} d \ldots \ldots \text { (ii) }$
Substituting the value of (ii) in (i), we get$\frac{3}{2} d+6 d=15$
$\Rightarrow 3 d+12 d=15 \times 2$
$\Rightarrow 15 d=30$
$\Rightarrow d=2$
$\Rightarrow a=\frac{3}{2} \times 2[\text { From (ii) }]$
$\Rightarrow a=3$
Thus, the A.P. is $3,5,7,9, \ldots .$.
View full question & answer→Question 663 Marks
Find the sum of,
All $3$ - digit natural numbers, which are multiples of $11.$
AnswerWe know that the first $3$ digit number multiples of $11$ will be $110.$
Last 3 digit number multiple of $11$ will be $990.$
So here,
First term $(a) = 110$
Last term $(l) = 990$
Common difference $(d) = 11$
So, here the first step is to find total number of terms.Let us take the number of terms as n.
Now, as we know,
$a_n = a + (n - 1)d$
So, for the last term,
$990 = 110 + (n - 1)11$
$990 = 110 + 11n - 11$
$990 = 99 + 11n$
$891 = 11n$
$81 = n$
Now, using the formula for the sum of terms, we get
$\text{S}_{\text{n}}=\frac{81}{2}[2(110)+(81-1)11]$
$\text{S}_{81}=\frac{81}{2}[220+80\times11]$
$\text{S}_{81}=\frac{81}{2}\times1100$
$\text{S}_\text{n}=81\times550$
$\text{S}_{81}=44550$
Therefore, the sum of all the $3$ digit multiples of $11$ is $44550.$
View full question & answer→Question 673 Marks
The first and the last term of an A.P. are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum?
Answer$a = 17$ and$ l = 350, d = 9$
Let number of term of the A.P. $= n$
$a_n = a + (n - 1)d$
$350 = 17 + (n - 1) \times 9 = 17 + 9n - 9$
$350 = 8 + 9n$
$9n = 350 - 8 = 342$
$\therefore\ \text{n}=\frac{342}{9}=38$
$\therefore$ Number of terms = $38$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{38}{2}[2\times17+37\times9]=19[34 + 333]$
$=19\times367=6973$
View full question & answer→Question 683 Marks
Find the sum of first 20 terms of the sequence whose $n^{\text {th }}$ term is $a_n=A n+B$.
Answer$n^{\text {th }} \text { term }=a_n=A n+B \text { and number of terms }=20$
$\therefore a_1=A \times 1+B=A+B$
$a_2=A \times 2+B=2 A+B$
$a_3=A \times 3+B=3 A+B$
$\therefore \text { First term (a) }=A+B$
$\text { Common difference }=a_2-a_1$
$=2 A+B-A-B=A$
$\therefore S_n=\frac{n}{2}[2 a+(n-1 d]$
$S_{20}=\frac{20}{2}[2(A+B)+(20-1) A]$
$=10[2 A+2 B+19 A]$
$=10[21 A+2 B]=210 A+20 B$
View full question & answer→Question 693 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$a^n = (-1)^n 2^n.$
Answer$a^n=(-1)^n 2^n$
Here, the $n ^{\text {th }}$ term is given by the above expression. So, to find the first term we use $n =1$, we get,
$a_1=(-1)^1 \cdot 2^1$
$=(-1) \cdot 2$
$=-2$
Similarly, we find the other four terms,
$\text { Second term }(n=2) \text {, }$
$a_2=(-1)^2 \cdot 2^2$
$=1 \cdot 4$
$=4$
$\text { Third term }(n=3) \text {, }$
$a_3=(-1)^3 \cdot 2^3$
$=(-1) \cdot 8$
$=-8$
Fourth term $(n=4)$,
$a_4=(-1)^4 \cdot 2^4$
$=1.16$
$=16$
Fifth term ( $n =5$ ),
$a_5=(-1)^5 \cdot 2^5$
$=(-1) \cdot 32$
$=-32$
Therefore, the first five terms of the given A.P are $a_1=-2, a_2=4, a_3=-8, a_4=16, a_5=-32$.
View full question & answer→Question 703 Marks
Write time first five terms of the following sequances whose $n ^{\text {th }}$ terms are:
$a_n=3 n+2$
Answer$a_n=3 n+2$
Let $n=1,2,3,4,5$, them
First five terms,
$a_1=3 \times 1+2=3+2=5$
$a_2=3 \times 2+2=6+2=8$
$a_3=3 \times 3+2=9+2=11$
$a_4=3 \times 4+2=12+2=14$
$a_5=3 \times 5+2=15+2=17$
View full question & answer→Question 713 Marks
The sum of three terms of an A.P. is $21$ and the product of the first and the third terms exceeds the second term by $6$, find three terms.
AnswerGiven,
Sum of three terms of on A.P. is $21.$
Product of first and the third term exceeds the secomd term by $6.$
Let, the three numbers be $a - d, a, a + d,$ with common differnce d: then,
$(a - d) + a + (a + d) = 21$
$3a = 21$
$\text{a}=\frac{21}{3}=7$
and $(a - d)(a + d) = a + 6$
$a^2 - d^2 = a + 6$
Put $a = 7 \Rightarrow 7^2 - d^2 = 7 + 6$
$49 - 13 = d^2$
$d^2 = 36$
$\text{d}=\sqrt{36}$
$\text{d}=\pm6$
$\therefore$ The three terms are $a - d, a, a + d$, i.e., $1, 7, 13.$
View full question & answer→Question 723 Marks
In which of the following situations, the sequence of numbers formed will form an A.P.?
The amount of air present in the cylinder when a vacuum pump removes each time $\frac{1}{4}$ of the remaining in the cylinder.
AnswerHere, let us take initial amount of air present in the cylinder as $100$ units.
Amount left after vacuum pump removes air for $1^{st}$ time $=100-\Big(\frac{1}{4}\Big)100$
$= 100 - 25 = 75$
Amount left after vacuum pump removes air for $2^{nd}$ time $=75-\Big(\frac{1}{4}\Big)75$
$= 75 - 18.75 = 56.25$
Amount left after vacuum pump removes air fir $3^{rd}$ time $=56.25-\Big(\frac{1}{4}\Big)56.25$
$= 56.25 - 14.06 = 42.19$
Thus, the amount left in the cylinder at various stages is $100, 75, 56.25, 42.19, .....$
Now, for a sequence to be an A.P., the difference between adjecent terms should be equal.
Here,
$a_1 - a = 75 - 100 = -25$
Also,
$a_2 - a_1 = 56.25 - 75 = -18.75$
Since, $\text{a}_2-\text{a}\neq\text{a}_2-\text{a}_1$
The sequence is not an A.P.
View full question & answer→Question 733 Marks
The first term of an A.P. is $5$, the common difference is $3$ and the last term is $80$?
AnswerThe First term of an A.P. $(a) = 5$
Common difference $(d) = 3$
Last term $=80$
Let the last term be $n^{\text {th }}$
$a_n=a+(n-1) d$
$\Rightarrow 80=5+(n-1) \times 3$
$\Rightarrow 80=5+3 n-3$
$\Rightarrow 3 n=80-5+3=78$
$\Rightarrow n=26$
Number of terms $=26$
View full question & answer→Question 743 Marks
Find the sum of,
All $2$ - digit natural numbers divisible by $4$.
AnswerWe can see it forms an A.P. as the common difference is $4$ and the first term is $4$ . To find no. of terms $n$,
We know that
$a_n=a+(n-1) d$
$96=12+(n-1) 4$
$84=(n-1) 4$
$21=n-1$
$22=n$
Now,
First term $(a) = 12$
Number of terms $( n ) =22$
Common difference $( d )=4$
Now, using the formula for the sum of $n$ terms, we get
$S_{22}=\frac{22}{2}\{2(12)+(22-1) 4\}$
$S_{22}=11\{24+84\}$
$S_{22}=1188$
Hence, The sum of $22$ terms is $1188$ which are divisible by $4$ .
View full question & answer→Question 753 Marks
For the following arithmetic progressions write the first term a and the common difference d:
$-1.1, -3.1, -5.1, -7.1, .....$
Answer$-1.1, -3.1, -5.1, -7.1, .....$
Here first term $(a) = -1.1$
We have $= a_2 - a_1 = -3.1 - (-1.1)$
$= -3.1 + 1.1 = -2.0$
$a_3 - a_2 = -5.1 - (-3.1) = -5.1 + 3.1 = -2.0$
$a_4 - a_3 = -7.1 - (-5.1) = -7.1 + 5.1 = -2.0$
$\therefore$ Common difference $= -2.0.$
View full question & answer→Question 763 Marks
In which of the following situations, the sequence of numbers formed will form an A.P.?
The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 20 for each succeeding metre.
AnswerCost of digging a well for the first metre = Rs. 150
Cost for the second metre = Rs. 150 + Rs. 20 = Rs. 170
Cost for the third metre = Rs. 170 + Rs. 20 = Rs. 190
Cost for the fourth metre = Rs. 190 + Rs. 20 = Rs. 210
The sequence will be (in rupees)
150, 170, 190, 210, .....
Which is an A.P.
Whose = 150 and d = 20.
View full question & answer→Question 773 Marks
Write the expression of the common difference of an A.P. whose first term is a and $n^{th}$ term is b.
AnswerHere, we are given
First term $= a$
Last term $= b$
Let us take the common difference as $d$
Now, we know
$a_n = a + (n - 1)d$
So,
For the last term $(a_n),$
$b = a + (n - 1)d$
$b - a = (n - 1)d$
$\text{d}=\frac{\text{b}-\text{a}}{\text{n}-1}$
Therefore, common difference of the A.P. is $\text{d}=\frac{\text{b}-\text{a}}{\text{n}-1}$.
View full question & answer→Question 783 Marks
For what value of n , the $n ^{\text {th }}$ terms of the arithmetic progressions $63,65,67, \ldots$ and $3,10,17, \ldots$ are equal?
AnswerIn the A.P. $63,65,67, \ldots .$.
$a=63$ and $d=65-63=2$
$a_n=a_1+(n-1) d=63+(n-1) \times 2=63+2 n-2=61+2 n$
and in the A.P. 3, 10, 17, .....
$a=3$ and $d=10-3=7$
$a_n=a+(n-1) d=3+(n-1) \times 7=3+7 n-7=7 n-4$
But both $n^{\text {th }}$ terms are equal
$61+2 n=7 n-4$
$\Rightarrow 61+4=7 n-2 n$
$\Rightarrow 65=5 n$
$\Rightarrow n =13$
$n =12$.
View full question & answer→Question 793 Marks
The third term of an A.P. is $7$ and the seventh term exceeds three times the third term by $2$. Find the first term, the common difference and the sum of first $20$ terms.
AnswerGiven, $a_3 = 7$ and $3a_3 + 2 = a_7 a_7 = 3 \times 7 + 2 a_7 = 21 + 2 = 23$
$\therefore$ $a_n = a + (n - 1)d a_3 = a + (3 - 1)d$ and
$a_7 = a + (7 - 1)d 7 = a + 2d .....(i) 23 = a + 6d .....(ii)$ Subtract $(i)$ from $(ii)$
$d = 4$ Put $d = 4 in (i) $
$\Rightarrow 7 = a + 2.4 a = 7 - 8 = -1$
Given to find sum of first $20$ terms. $\text{S}_{20}=\frac{20}{2}\big[-2+(20-1)4\big]$
$=10(-2+76)$ $\therefore\ \text{S}_{20}=740$ View full question & answer→Question 803 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{3\text{n}-2}{5}$
Answer$\text{a}_\text{n}=\frac{3\text{n}-2}{5}$
Let $n = 1, 2, 3, 4, 5$, then
$\text{a}_1=\frac{3\times1-2}{5}=\frac{3-2}{5}=\frac{1}{5}$
$\text{a}_2=\frac{3\times2-2}{5}=\frac{6-2}{5}=\frac{4}{5}$
$\text{a}_3=\frac{3\times3-2}{5}=\frac{9-2}{5}=\frac{7}{5}$
$\text{a}_4=\frac{3\times4-2}{5}=\frac{12-2}{5}=\frac{10}{5}=2$
$\text{a}_5=\frac{3\times5-2}{5}=\frac{15-2}{5}=\frac{13}{5}$
View full question & answer→Question 813 Marks
Write the sequence with $n^{th}$ term:
$a_n = 6 - n.$
Show the all of the above sequences form A.P.
Answer$a_n=6-n$
Now,to show that it is an A.P, we will find its few terms by substituting $n =1,2,3$
So, Substituting $n=1$, we get $a_1=6-1 a_1=5$
Substituting $n=2$, we get $a_2=6-2 a_2=4$
Substituting $n=3$, we get $a_3=6-3 a_3=3$
Further, for the given to sequence to be an A.p.
Common difference $(d)=a_2-a_1=a_3-a_2$
Here, $a_2-a_1=4-5=-1$
Also, $a_3-a_2=3-4=-1$
Since $a_2-a_1=a_3-a_2$
Hence, the given sequence is an A.P.
View full question & answer→Question 823 Marks
A piece of equipment cost a certain factory Rs. 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
AnswerCost of a piece of equipment = Rs. 600,000Rate of depreciation for the first year = 15%
For the second year = 13.5%
For the third year = 12.0% and so on
The depreciation is in A.P.
Whose first term (a) = 15
and common difference (d) = 13.5 - 15.0 = -1.5
Period (n) = 10
$\therefore$ Total depreciation % $(\text{S}_\text{n})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{10}{2}[2\times15+(10-1)(-1.5)]$
$=5[30+9\times(-1.5)]$
$=5[30-13.5]=5\times16.5=82.5$
$\therefore$ Total depreciation $=\frac{600000\times82.5}{100}$
= Rs. 495000
$\therefore$ Its value at the end of 10 years
= Rs. 600000 - Rs. 495000 = Rs. 105000
View full question & answer→Question 833 Marks
Find the sum of the following arithmetic progressions:
$a + b, a - b, a - ab$, ..... to $22$ terms.
AnswerIn an A.P. let first term $=a$, common difference $= d$, and there are n terms. Then, sum of n terms is,
$S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$
Given,
$a+b, a-b, a-3 b, \ldots . .$
Here,
First term, $a=a+b$,
Difference $d=a-b-(a+b)=a-b-a-b=-2 b$
and no of terms $n =22$
We know $S _{ n }=\frac{ n }{2}\{2 a +( n -1) d \}$
$\Rightarrow S_n=\frac{22}{2}[2(a+b)+(22-1)-2 b]$
$\Rightarrow S_n=11[2 a+2 b+21 \times-2 b]$
$\Rightarrow S_n=11[2 a+2 b-42 b]$
$\Rightarrow S_n=11[2 a-40 b]$
$\Rightarrow S_n=22 a-440 b$
Hence, Sum of $22$ terms is $22$ $a - 440b.$
View full question & answer→Question 843 Marks
If $S_n$ denotes the sum of the first n terms of an A.P., prove that $S_{30} = 3(S_{20} - S_{10}).$
AnswerSn is the sum of first n terms of an A.P.
To prove: $S_{30} = 3(S_{20} - S_{10})$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{R.H.S.} = 3(\text{S}_{20} - \text{S}_{10})$
$=3\Big[\frac{20}{2}\big(2\text{a}+(20-1)\text{d}\big)-\frac{10}{2}\big(2\text{a}+(10-1)\text{d}\Big]$
$=3\Big[10\big(2\text{a}+19\text{d}\big)-5\big(2\text{a}+9\text{d}\big)\Big]$
$3\Big[20\text{a}+190\text{d}-10\text{a}-45\text{d}\Big]$
$=3\big[10\text{a}+145\text{d}\big]$
$30\text{a}+435\text{d}$
$15\big[2\text{a}+29\text{d}\big]$
$=\frac{30}{2}\big[2\text{a}+(30-1)\text{d}\big]$
$=\text{S}_{30}=\text{R.H.S.}$
View full question & answer→Question 853 Marks
Find the sum of,
The first 40 positive integers divisible by,
- 3
- 5
- 6
Answer
- First 40 positive integers divisible by 3
are 3, 6, 9, 12, 15, ....., 120
in which a = 3, d = 3 and l = 120
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{40}=\frac{40}{2}[2\times3+(40-1)\times3]$
= 20[6 + 39 × 3]
= 20[6 + 117] = 20 × 123 = 2460
- 40 multiple of 5 are
In which first term (a) = 5
Common difference (d) = 5
and last term (l) = 200
$\therefore\ \text{S}_{\text{n}}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{40}=\frac{40}{2}[2\times5+(40 - 1)\times5]$
= 20[10 + 39 × 5] = 20[10 + 195]
= 20 × 205 = 4100
- 40 multiples of 6 are
6, 12, 18, 24, ....., 240
In which first term (a) = 6
Common difference (d) = 6
$\therefore\ \text{S}_{\text{n}}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{40}=\frac{40}{2}[2\times6+(40 - 1)\times6]$
= 20[12 + 39 × 6] = 20[12 + 234]
= 20 × 246 = 4920 View full question & answer→Question 863 Marks
Find the indicated terms in the following sequences whose $n^{th}$ terms are:
$a_n = n(n - 1) (n - 2); a_5$ and $a_8$.
AnswerGiven $n^{\text {th }}$ term is $a_n=n(n-1)(n-2)$
To find $5^{\text {th }}, 8^{\text {th }}$ terms of given sequence, put $n =5,8$ an then, we get
$a_5=5(5-1)(5-2)=5 \times 4 \times 3=60 \\
a_8=8(8-1)(8-2)=8 \times 7 \times 6=336$
$\therefore$ The required terms are $a_5=60$ and $a_8=336$.
View full question & answer→Question 873 Marks
In an A.P., if the first term is $22$, the common difference is $-4$ and the sum to n terms is $64$, find n.
AnswerGiven,
$a = 22, d = -4, S_n = 64$
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$64=\frac{\text{n}}{2}\big[2\times22+(\text{n}-1)(-4)\big]$
$64=\text{n}(24-2\text{n})$
$64=2\text{n}(12-\text{n})$
$12\text{n}-\text{n}^2=\frac{64}{2}=32$
$\text{n}^2-12\text{n}+32=0$
$(\text{n}-4)(\text{n}-8)=0$
$\therefore\ \text{n}=4\text{ or }8$
View full question & answer→Question 883 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8}, .....$
AnswerIn the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference (d).
$\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8}, .....$
in the given sequance
$\text{a}_1=\frac{1}{2},\text{a}_2=\frac{1}{4},\text{a}_3=\frac{1}{6},\text{a}_4=\frac{1}{8}.$
Check the condition
$a_2 - a_1 = a_3 - a_2$
$\frac{1}{4}-\frac{1}{2}=\frac{1}{6}-\frac{1}{4}$
$\frac{1-2}{4}=\frac{4-6}{24}$
$\frac{-1}{4}=-\frac{2}{24}$
$\frac{-1}{4}\neq\frac{-1}{12}$
Condition is not sarisfied
$\therefore$ Given sequnce not in A.P.
View full question & answer→Question 893 Marks
Write the sequence with $n^{th}$ term:
$a_n = 5 + 2n.$
Show the all of the above sequences form A.P.
Answer$a_n = 5 + 2n$
Put $n = 1, 2, 3...$
$a_1 = 5 + 2 \times 1 = 7$
$a_2 = 5 + 2 \times 2 = 9$
$a_3 = 5 + 2 \times 3 = 11$
$a_2 - a_1 = a_3 - a_2$
$9 - 7 = 11 - 9$
$2 = 2$
Common difference is $2$
So that
A.P, $7, 9, 11, 13, 15...$
View full question & answer→Question 903 Marks
Find:
$11^{th}$ term of the A.P. $10.0, 10.5, 11.0, 11.5, .....$
AnswerGiven,
A.P. $10.0, 10.5, 11.0, 11.5, .....$
Here,
First term $a = 10.0$
Common difference $d = 10.5 - 10.0 = 0.5$
We have to find $10^{th}$ term,
So Putting $n = 11$
We know, $n^{th}$ term of A.P.
$a_n = a + (n - 1)d$
$\Rightarrow a_{11} = 10.0 + (11 - 1)0.5$
$\Rightarrow a_{11} = 10 + 10 \times 0.5$
$\Rightarrow a_{11} = 10 + 5$
$\Rightarrow a_{11} = 15$
Hence, $11^{th}$ term of given A.P. is $15$
View full question & answer→Question 913 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$3, 6, 12, 24, .....$
AnswerIn the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference $(d)$.
$3, 6, 12, 24, .....$
Here,
First term $(a) = 3$
$a_1 = 6$
$a_2 = 12$
Now, for the given to sequance to be an A.P,
Common difference $(d) = a_1 - a = a_2 - a_1$
Here,
$a_1 - a = 6 - 3$
$= 3$
Also,
$a_2 - a_1 = 12 - 6$
$= 6$
Since $\text{a}_1-\text{a}\neq\text{a}_2-\text{a}_1$
Hence, the given sequence is not an A.P.
View full question & answer→Question 923 Marks
Find the indicated terms in the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{3\text{n}-2}{4\text{n}+5};\text{a}_7\text{ and }\text{a}_8$
Answer$a_{n}=\frac{3 n-2}{4 n+5}$
We need to find $a_7$ and $a_8$
Now, to find $a_7$ term we use $n=7$, we get,
$a _7=\frac{3(7)-2}{4(7)+5}$
$=\frac{21-2}{28+5}$
$=\frac{19}{33}$
Also, to find $a_8$ term we use $n=8$, we get
$a _8=\frac{3(8)-2}{4(8)+5}$
$=\frac{24-2}{32+5}$
$=\frac{22}{37}$
Thus, $a _7=\frac{19}{33}$ and $a _8=\frac{22}{37}$
View full question & answer→Question 933 Marks
Find:
Which term in the A.P. $84,80,76, \ldots .$. is $0$ ?
AnswerIn the given problem, we are given an A.P. and the value of one of its term. We need to find which term it is $( n )$.
So here we will find the value of $n$ using the formula, $a_n=a+(n-1) d$.
Given,
A.P., $84, 80, 76, .....$
$a_n=0$
Here,
First term $= 84$
Difference $=(80-84)=-4$
We have to find which term of A.P. is $0$
We know, $n ^{\text {th }}$ term of A.P.
$a_n=a+(n-1) d$
$\Rightarrow 0=84+(n-1)-4$
$\Rightarrow 0=84+(-4 n+4)$
$\Rightarrow 0=84-4 n+4$
$\Rightarrow 4 n=88$
$\Rightarrow n=\frac{88}{4}$
$\Rightarrow n=22$
Hence, $22^{\text {th }}$ term of the given A.P. is $0$ .
View full question & answer→Question 943 Marks
Find the sum of all integers between $50$ and $500$, which are divisible by $7$.
AnswerAll integers between $50$ and $500$ which are divisible by $7$, are
$56, 63, 70, 77, ....., 497$
($\because$ $497$ is divisible by $7$)
In which $a = 56, d = 63 - 56 = 7$ and $l = 497$
$a_n = a + (n - 1)d$
$\Rightarrow 497 = 56 + (n - 1)7 $
$\Rightarrow 497 = 56 + 7n - 7$
$\Rightarrow 7n = 497 - 56 + 7 = 448$
$\Rightarrow\text{n}=\frac{448}{7}=64$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]=\frac{64}{2}[56+497]$
$=32(553)=17696$
View full question & answer→Question 953 Marks
Find the sum of the following arithmetic progressions:
$3,\frac{9}{2},6,\frac{15}{2}, .....,\text{to }12\text{ terms}.$
AnswerIn an A.P. let first term = $a$, common difference = $d$, and there are $n$ terms. Then, sum of $n$ terms is,
$\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
Common difference of the A.P. $(d) = a_2 - a_1$
$=\frac{9}{2}-3$
$=\frac{9-6}{2}$
$=\frac{3}{2}$
Number of terms $(n) = 25$
First term for the given A.P. $(a) = 3$
So, using the formula we get,
$\text{S}_{25}=\frac{25}{2}\Big[2(3)+(25-1)\Big(\frac{3}{2}\Big)\Big]$
$=\Big(\frac{25}{2}\Big)\Big[6+(24)\Big(\frac{3}{2}\Big)\Big]$
$=\Big(\frac{25}{2}\Big)\Big[6+\Big(\frac{72}{2}\Big)\Big]$
$=\Big(\frac{25}{2}\Big)[6+36]$
$=\Big(\frac{25}{2}\Big)[42]$
$=(25)(21)$
$=525$
On further simplifying, we get,
$S_{25} = 525$
Therefore, the sum of first $25$ terms for the given A.P. is $525.$
View full question & answer→Question 963 Marks
Find the common difference and write the next four terms of the following arithmetic progressions:
$-1,\frac{1}{4},\frac{3}{2},\ .....$
AnswerGiven arithmetic progression is,
$-1,\frac{1}{4},\frac{3}{2},\ .....$
$\text{a}_1=-1,\text{a}_2=\frac{1}{4},\text{a}_3=\frac{3}{2},\ .....$
Common difference $(d) = a_2 - a_1$
$=\frac{1}{4}-(-1)$
$=\frac{1+4}{4}$
$\text{d}=\frac{5}{4}$
To find next for terms.
$\text{a}_4=\text{a}_3+\text{d}=\frac{3}{2}+\frac{5}{4}=\frac{6+5}{4}=\frac{11}{4}$
$\text{a}_5=\text{a}_4+\text{d}=\frac{11}{4}+\frac{5}{4}=\frac{16}{4}$
$\text{a}_6=\text{a}_5+\text{d}=\frac{16}{4}+\frac{5}{4}=\frac{21}{4}$
$\text{a}_7=\text{a}_6+\text{d}=\frac{21}{4}+\frac{5}{4}=\frac{26}{4}$
$\therefore\ \text{d}=\frac{5}{4},\text{d}_4=\frac{11}{4},\text{a}_5=\frac{16}{4},\text{a}_6=\frac{21}{4},\text{a}_7=\frac{26}{4}.$
View full question & answer→Question 973 Marks
Find:
Is $302$ a term of the A.P. $3, 8, 13, .....?$
AnswerIn the given problem, we are given an A.P. and the Value of one of its term.
We need to find whether it is a term of the A.P. or not so here we will use the formula $a_n = a + (n - 1)d.$
Here,
A.P.$ = 3, 8, 13, .....$
First term $(a) = 3$
and Common difference $(d) = 8 - 3 = 5$
Let $a_n = 302$, then
$302 = a + (n - 1)d = 3 + (n - 1) \times 5$
$\Rightarrow 302 = 3 + 5n - 5$
$\Rightarrow 302 - 3 + 5 = 5n$
$\Rightarrow\ 304=5\text{n}\Rightarrow\text{n}=\frac{304}{5}=60\frac{4}{5}$
Since n is not a natural number
$\therefore$ $302$ is not a term of the given sequence.
View full question & answer→Question 983 Marks
Write the arithmetic progression when first term a and common difference d are as following:
a = -1.5, d = -0.5.
AnswerFirst term (a) = -1.5
and common difference (d) = -0.5
$\therefore$ Second term = a + d = -1.5 + (-0.5)
= -1.5 - 0.5 = -2.0
Third term = a + 2d = -1.5 + 2(-0.5)
= -1.5 - 1.0 = -2.5
Fourth term = a + 3d = -1.5 + 3(0.5)
= -1.5 - 1.5 = -3.0
Fifth term = a + 4d = -1.5 + 4(-0.5)
= -1.5 - 2.0 = -3.5
$\therefore$ AP Will be -1.5, -2.0, -2.5, -3.0, -3.5, .....
View full question & answer→Question 993 Marks
Find the number of terms of the A.P. $-12, -9, -6, ....., 21$. If $1$ is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.
AnswerFirst term $a_1 = -12$ Common difference,
$d = a_2 - a_1 = -9 - (-12) = 3 a_n = 21 $
$\Rightarrow a + (n - 1)d = 21 $
$\Rightarrow -12 + (n - 1) \times 3 = 21$
$\Rightarrow 3n = 36 $
$\Rightarrow n = 12$
Therefore, number of terms in the given A.P. is $12$.
Now, when 1 is added to each of the $12$ terms, the sum will increase by $12$.
So, the sum of all terms of the A.P. thus obtained $= S_{12} + 12$
$=\frac{12}{2}[2(-12)+11(3)]+12$ $= 6 \times (9) + 12 = 66$
View full question & answer→Question 1003 Marks
Which term of the sequence $114, 109, 104, .....$ is the first negative term?
AnswerHere,
$A.P. is 114, 109, 104, .....$
So, first term $a = 114$
Now,
Common difference $(d) = a_1 - a$
$= 109 - 114$
$= -5$
Now, we need to find the first negative term,
$a_n < 0$
$114 + (n - 1)(-5) < 0$
$114 - 5n + 5 < 0$
$119 - 5n < 0$
$5n > 119$
Further simplifying, we get,
$\text{n}>\frac{119}{5}$
$\text{n}>23\frac{4}{5}$
$\text{n}\geq24$ (as n is a natural number)
Thus, $n = 24$
Therefore, the first negative term is the $24^{th}$ term of the given A.P.
View full question & answer→Question 1013 Marks
The angles of a quadrilateral are in A.P. whose common difference is 10º. Find the numbers.
AnswerA quadrilateral has four angles. Given, four angles are in A.P. with common difference 10.
Let the four angles be, a - 3d, a - d, a + d, a + 3d with common difference = 2d.
2d = 10
$\text{d}=\frac{10}{2}=5$
In a quadrilateral, sum of all angles = 360º
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 360
4a = 360
$\text{a}=\frac{\text{360}}{4}=90^\circ$
$\therefore$ The angles are a - 3d, a - d, a + d, a + 3d with a = 90, d = 5
i.e. 90 - 3(5), 90 - 5, 90 + 3(5)
⇒ 75º, 85º, 95º, 105º.
View full question & answer→Question 1023 Marks
The $n^{th}$ term of an A.P is given by $(-4n + 15)$, Find the sum of first $20$ terms of this A.P.
Answer$T_n = (-4n + 15), S_{20} = ?$
$T_1 = -4 \times 1 + 15 = -4 + 15 = 11$
$T_2 = -4 \times 2 + 15 = -8 + 15 = 7$
$\therefore$ $a = 11, d = T_2 - T_1 = 7 - 11 = -4$
$\text{S}_{20}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{20}{2}[2\times11+(20-1)(-4)]$
$=10[22+19(-4)]$
$=10(22-76)=10\times(-54)=-540$
View full question & answer→Question 1033 Marks
How many terms of the A.P. $9, 17, 25, .....$ must be taken so that sum is $636$?
AnswerGiven,
A.P. $9,17,25$,
$a=9, d=17-9=8$, and $S_n=636$
$636=\frac{ n }{2}(2.9+( n -1) 8)$
$1272=n(18-8+8 n)$
$1272=n(10+8 n)$
$2 \times 636=2 n(5+4 n)$
$636=5 n+4 n^2$
$4 n^2+5 n-636=0$
$4 n^2+53 n-48 n-636=0$
$4 n^2-48 n+53 n-636=0$
$4 n(n-12)+53(n-12)=0$
$(4 n+53)(n-12)=0$
$\therefore n =12$ (Since $n \frac{-53}{4}$ is not a natural number)
Therefore, value of $n$ is $12$ .
View full question & answer→Question 1043 Marks
Write the sum of first n even natural numbers.
AnswerLet,
Even numbers are, 2, 4, 6, 8, .....
Here,
First term a = 2
Difference d = 4 - 2 = 2
We know. Sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(2)+(\text{n}-1)2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[4+2\text{n}-2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{n}+2]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2[\text{n}+1]$
$\Rightarrow\ \text{S}_\text{n}=\text{n}(\text{n}+1)$
Hence, Sum of even numbers is n(n + 1).
View full question & answer→Question 1053 Marks
Write the arithmetic progression when first term a and common difference d are as following:
$\text{a}=-1,\text{d}=\frac{1}{2}$
Answer$\text{a}=-1,\text{d}=\frac{1}{2}$
Now, as $a = -1$
A.P. would be represented by $a, a_1, a_2, a_3, a_4, .....$
So,
$a_1 = a + d$
$\text{a}_1=-1+\Big(\frac{1}{2}\Big)$
$\text{a}_1=\frac{-2+1}{2}$
$\text{a}_1=\frac{-1}{2}$
Similarly.
$a_2 = a_1 + d$
$\text{a}_2=\frac{-1}{2}+\Big(\frac{1}{2}\Big)$
$a_2 = 0$
Also,
$a_3 = a_2 + d$
$\text{a}_3=0+\Big(\frac{1}{2}\Big)$
$\text{a}_3=\frac{1}{2}$
Further,
$a_4 = a_3 + d$
$\text{a}_4=\Big(\frac{1}{2}\Big)+\Big(\frac{1}{2}\Big)$
$\text{a}_4=\frac{2}{2}$
$\text{a}_4=1$
Therefore, A.P. with $a = -1$ and $\text{d}=\frac{1}{2}$ is $-1,\frac{-1}{2},0,\frac{1}{2},1\ .....$
View full question & answer→Question 1063 Marks
Find n if the given value of $x$ is the $n^{th}$ term of the given A.P.
$-1, -3, -5, -7, .....; x = -151.$
AnswerGiven sequence $-1, -3, -5, -7, .....; x = -151$
First term $(a) = -1$
Common difference $(d) = -3 - (-1) = -3 + 1 = -2$
$n^{th}$ term $a_n = a + (n - 1)d$
Given $a_n = -151,$
$-151 = -1 + (n - 1)(-2)$
$-151 + 1 = (n - 1)(-2)$
$-150 = (n - 1)(-2)$
$=\frac{-150}{-2}=\text{n}-1$
$75 = n - 1$
$75 + 1 = n$
$n = 76.$
View full question & answer→Question 1073 Marks
If the sum of a certain number of terms starting from first term of an A.P. is $25, 22, 19, ...,$ is $116$. Find the last term.
AnswerGiven A.P. is $25, 22, 19, .....$
First term $(a) = 25, d = 22 - 25 = -3.$
Given, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$116=\frac{\text{n}}{2}\big[2\times25+(\text{n}-1)(-3)\big]$
$232 = n(50 - 3(n - 1))$
$232 = n(53 - 3n)$
$232 = 53n - 3n^2$
$3n^2 - 53n + 232 = 0$
$3n^2 - 29n - 24n + 232 = 0$
$3n^2 - 24n - 29n + 232 = 0$
$3n(n - 8) - 29(n - 8) = 0$
$(n - 8)(3n - 29)$
$\therefore$ $n = 8$
$\Rightarrow a_8 = 25 + (8 - 1)(-3)$
$\therefore$ $n = 8, a_8 = 4$
$= 25 - 21 = 4.$
View full question & answer→Question 1083 Marks
Which term of the A.P. $3,15,27,39, \ldots$ will be $120$ more than its $21^{\text {st }}$ term?
AnswerGiven,
A.P. is $3,15,27,39, \ldots .$.
Let $n^{\text {th }}$ term is $120$ more than $21^{\text {st }}$ term
Then $a_n=120+a_{21}$
For the given sequence
$a=3, d=15-3=12$
$a+(n-1) d=120+a+(21-1) d$
$(n-1) 12=120+20(12)$
$(n-1) 12=360$
$(n-1)=\frac{360}{12}=30$
$n=31$
$\therefore 31^{\text {st }}$ term is $120$ more than $21^{\text {st }}$ term.
View full question & answer→Question 1093 Marks
Find:
Which term of the A.P.$ 3, 8, 13, .....$ is $248$?
AnswerIn the given problem, we are given an A.P. and the value of one of its term. We need to find which term it is ( n ). So here we will find the value of $n$ using the formula, $a_n=a+(n-1) d$.
Here,
A.P. is $3,8,13, \ldots .$.
$a_n=248$
$a=3$
Now,
$\text { Common difference }(d)=a_1-a$
$=8-3$
$=5$
Thus, using the above mentioned formula
$a_n=a+(n-1) d$
$248=3+(n-1) 5$
$248-3=5 n-5$
$245+5=5 n$
$n=\frac{250}{5}$
$n=50$
Thus, $n =50$
Therefore 248 is the $50^{\text {th }}$ term of the given A.P.
View full question & answer→Question 1103 Marks
The first term of an A.P. is $5$, the last term is $45$ and the sum is $400$. Find the number of terms and the common difference.
AnswerGiven,
$a = 5, l = 45$, Sum of terms $= 400$
$\therefore\ \text{S}_{\text{n}}=400$
$\frac{\text{n}}{2}\{5+45\}=400$
$\frac{\text{n}}{2}=50=400$
$\text{n}=40\times\frac{2}{5}$
$\therefore\ \text{n}=16$
$16^{th}$ term is 45
$\text{a}_{16}=45\Rightarrow5+(16-1)\text{d}=45\Rightarrow15\text{d}=40$
$\text{d}=\frac{40}{15}=\frac{8}{3}$
$\therefore\ \text{n}=16,\text{d}=\frac{8}{3}$
View full question & answer→Question 1113 Marks
Find n if the given value of $x$ is the $n^{th}$ term of the given A.P.
$5\frac{1}{2},11,16\frac{1}{2},22,\ .....;\text{ x}=550$
AnswerWe are given,
$a_n = 550$
Let us take the total number of terms as $n$,
So,
First term $(\text{a})=5\frac{1}{2}$
Last term $(a_n) = 550$
Common difference $(\text{d})=11-5\frac{1}{2}$
$=11-\frac{11}{2}$
$=\frac{22-11}{2}$
$=\frac{11}{2}$
Now, as we know,
$a_n = a + (n - 1)d$
So, for the last term.
$550=5\frac{1}{2}+(\text{n}-1)\Big(\frac{11}{2}\Big)$
$550=\frac{11}{2}+\frac{11}{2}\text{n}-\frac{11}{2}$
$550=\frac{11}{2}\text{n}$
$\text{n}=\frac{550(2)}{11}$
On further simplifying, we get,
$\text{n}=\frac{1100}{11}$
$n = 100$
Therefore, the total number of terms of the given A.P. is $n = 100.$
View full question & answer→Question 1123 Marks
The sums of first n terms of three A.P. S are $S_1, S_2$_ and $S_3.$ The first term of each is 5 and their common differences are $2, 4$ and $6$ respectively. Prove that $S_1 + S_3 = 2S_2.$
Answer$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Now, For $S_1$, when $a = 5$ and $d = 2$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2(1)+(\text{n}-1)(2)]=\frac{\text{n}}{2}[10+2\text{n}-2]$
$=\frac{\text{n}}{2}[2\text{n}+8]$
$\text{s}_1=\text{n}^2+4\text{n}\ ...(\text{i})$
For $S_2$, when $a = 5$ and $d = 4$
$\text{S}_2=\frac{\text{n}}{2}[2(5)+(\text{n}-1)(4)]$
$=\frac{\text{n}}{2}[10+4\text{n}-4]$
$=\frac{\text{n}}{2}[4\text{n}+6]$
$\text{S}_2=2\text{n}^2+3\text{n}\ ...(\text{ii})$
For $S_3$, when $a = 5$ and $d = 6$
$\text{S}_3=\frac{\text{n}}{2}[2(5)+(\text{n}-1)(6)]$
$=\frac{\text{n}}{2}[10+6\text{n}-6]=\frac{\text{n}}{2}[6\text{n}+4]$
$\text{S}_3=3\text{n}^2+2\text{n}\ ...(\text{iii})$
Putting the value of $S_1, S_2$ and $S_3$ in equation
$S_1 + S_3 = 2S_2$
From equation (i), (ii) and (iii)
$(n^2 + 4n) + (3n^2 + 2n) = 2(2n^2 + 3n)$
$4n^2 + 6n = 4n^2 + 6n$
$0 = 0$
Hence proved.
View full question & answer→Question 1133 Marks
All integers between $1$ and $500$ which are multiplies of $2$ as well as of $5$.
Answer$\therefore$ Multiples of 2 as well as of 5 between 1 and 500 is $10,20,30, \ldots . . . ., 490$
which from an AP with first term (a) $=10$ and common difference $( d )=20-10=10$
$n ^{\text {th }}$ term $a _{ n }=$ Last term $( l )=490$
$\therefore$ Sum of $n$ terms between 1 and 500
$S_n=\frac{n}{2}[a+l] \ldots \ldots(i)$
$\because a_n=a+(n-1) d=I$
$\Rightarrow 10+(n-1) 10=490$
$\Rightarrow(n-1) 10=480$
$\Rightarrow n-1=48$
$\Rightarrow n=49$
From Eq. (i), $S _{49}=\frac{49}{2}(10+490)$
$=\frac{49}{2} \times 500$
$=49 \times 250=12250$
View full question & answer→Question 1143 Marks
The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{th}$ term.
AnswerLet a and d be the first term and common difference of an A.P, respectively.
Now, by given condition, $\text{a}_8=\frac{1}{2}\text{a}_2$
$\Rightarrow\ \text{a}+7\text{d}=\frac{1}{2}(\text{a}+\text{d})\ [\because \text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d]}$
$\Rightarrow\ 2\text{a}+14\text{d}=\text{a}+\text{d}$
$\Rightarrow\ \text{a}+13\text{d}=0\ .....\text{(i)}$
and $\text{a}_{11}=\frac{1}{3}\text{a}_4+1$
$\Rightarrow\ \text{a}+10\text{d}=\frac{1}{3}[\text{a}+3\text{d}]+1$
$\Rightarrow\ 3\text{a}+30\text{d}=\text{a}+3\text{d}+3$
$\Rightarrow\ 2\text{a}+27\text{d}=3\ .....\text{(ii)}$
From Eqs. (i) and (ii),
$2(-13d) + 27d = 3$
$\Rightarrow -26d + 27d = 3$
$\Rightarrow d = 3$
From Eq. (i),
$a + 13(3) = 0 \Rightarrow a = -39$
$\therefore$ $a_{15} = a + 14d = -39 + 14(3)$
$= -39 + 42 = 3.$
View full question & answer→Question 1153 Marks
Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find.
n and $a _{ n }$, if $a =2, d=8$ and $S _{ n }=90$.
Answer$a = 2, d = 8, S_n = 90$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\Rightarrow90=\frac{\text{n}}{2}[2\times2+(\text{n}-1)8]$
$\Rightarrow 180 = n(4 + 8n - 8) \Rightarrow 180 = n(8n - 4)$
$\Rightarrow 8n^2 - 4n - 180 = 0$
$\Rightarrow 2n^2 - n - 45 = 0$ (Dividing by $4$)
$\Rightarrow 2n^2 - 10n + 9n - 45 = 0$
$\begin{Bmatrix}\because-45\times2=-90\\ \therefore-90=-10\times9\\ -1=-10+9 \end{Bmatrix}$
$\Rightarrow 2n (n - 5) + 9(n - 5) = 0$
$\Rightarrow (n - 5)(2n + 9) = 0$
Either $n - 5 = 0$, then $n = 5$
or $2n + 9 = 0$,
Then, $2\text{n}=-9\Rightarrow\ \text{n}=\frac{-9}{2}$ but is is not possible being fraction
$\therefore\ \text{n}=5$
$a_n = a + (n - 1)d = 2 + (5 - 1) \times 8$
$= 2 + 4 \times 8 = 2 + 32 = 34$
View full question & answer→Question 1163 Marks
The sum of first n terms of an A.P. is $3n^2 + 4n$. Find the $25^{th}$^ term of this A.P.
Answer$S_n = 3n^2 + 4n$
We know
$a_n = S_n - S_{n-1}$
$\therefore$$a_n = 3n^2 + 4n - 3(n - 1)^2 - 4(n - 1)$
$\Rightarrow a_n = 6n + 1$
$a_{25} = 6(25) + 1 = 151$
View full question & answer→Question 1173 Marks
If the sum of n terms of an A.P. is $S_n = 3n^2 + 5n$. Write its common difference.
AnswerHere, we are given,
$S_n = 3n^2 + 5n$
Let us take the first term as a and the common difference as d.
Now, as we know,
$a_n = S_n - S_{n-1}$
So, we get,
$a_n = (3n^2 + 5n) - [3(n - 1)^2 + 5(n - 1)]$
$= 3n^2 + 5n - [3(n^2 + 1 - 2n) + 5n - 5] [\text {Using}(a - b)^2 = a^2 + b^2 - 2ab]$
$= 3n^2 + 5n - (3n^2 + 3 - 6n + 5n - 5)$
$= 3n^2 + 5n - 3n^2 - 3 + 6n - 5n + 5$
$= 6n + 2 .....(i)$
Also,
$a_n = a + (n - 1)d$
$= a + nd - d$
$= nd + (a - d) .....(ii)$
On comparing the terms containing n in (i) and (ii), we get,
$dn = 6n$
$d = 6$
Therefore, the common difference is $d = 6.$
View full question & answer→Question 1183 Marks
For the following arithmetic progressions write the first term a and the common difference d:
$\frac{1}{5},\frac{3}{5},\frac{5}{5},\frac{7}{5}.$
Answer$\frac{1}{5},\frac{3}{5},\frac{5}{5},\frac{7}{5}.$
Here, first term of the given A.P. is $\text{(a)}=\frac{1}{5}$
Now, we will find the difference between the two terms of the given A.P.
$\text{a}_2-\text{a}_1=\frac{3}{5}-\frac{1}{5}$
$\text{a}_2-\text{a}_1=\frac{2}{5}$
Similarly,
$\text{a}_3-\text{a}_2=\frac{5}{5}-\frac{3}{5}$
$\text{a}_3-\text{a}_2=\frac{2}{5}$
Also,
$\text{a}_4-\text{a}_3=\frac{7}{5}-\frac{5}{5}$
$\text{a}_4-\text{a}_3=\frac{2}{5}$
As $\text{a}_2-\text{a}_1=\text{a}_3-\text{a}_2=\text{a}_4-\text{a}_3=\frac{2}{5}$
Therefore, the first term of the given A.P. is $\text{a}=\frac{1}{5}$ and the common differece is $\text{d}=\frac{2}{5}$.
View full question & answer→Question 1193 Marks
If the $5^{th}$ term of an A.P. is $31$ and $25^{th}$ term is $140$ more than the $5^{th}$ term, find the A.P.
AnswerWe know that,
$T_n = a + (n - 1)d$
$T_5 = a + 4d$
$\Rightarrow a + 4d = 31 .....(i)$
and $T_{25} = a + 24d$
$\Rightarrow a + 24d = 140 + T_5$
$\Rightarrow a + 24d = 140 + 31 = 171 .....(ii)$
Subtracting (i) from (ii).
$20d = 140$
$\text{d}=\frac{140}{20}=7$
and $a + 4d = 31$
$\Rightarrow a + 4 \times 7 = 31$
$\Rightarrow a + 28 = 31$
$\Rightarrow a + 31 - 28 = 3$
$a = 3$ and $d = 7$
A.P. will be $3, 10, 17, 24, 31, .....$
View full question & answer→Question 1203 Marks
The sum of first $q$ terms of an A.P. is $63 q-3 q^2$. If its $p^{\text {th }}$ term is -60 , find the value of $p$. Also, find the $11^{\text {th }}$ term of this A.P.
Answer$S_q=63 q-3 q^2$
We know
$a_q=S_q-S_{q-1}$
$\therefore a_q=63 q-3 q^2-63(q-1)+3(q-1)^2$
$a_q=66-6 q$
Now, $a_p=-60$
$\Rightarrow 60-6 p=-60$
$\Rightarrow 126=6 p$
$\Rightarrow p=21$
$a_{11}=66-6 \times 11=0$
View full question & answer→Question 1213 Marks
How many terms are there in the A.P.?
$18, 15\frac{1}{2},13,\ ....,-47$
AnswerGiven,
A.P. $18, 15\frac{1}{2},13,\ ....,-47$
Here,
First term $a = 18$
Difference $\text{d}=15\frac{1}{2}-18=\frac{-5}{2}$
Last $n^{th}$ term $a_n = -47$
We know, $n^{th}$ term of A.P.
$a_n = a + (n - 1)d$
$\Rightarrow\ -47=18+(\text{n}-1)\frac{-5}{2}$
$\Rightarrow\ -47=18\frac{-5\text{n}}{2}+\frac{5}{2}$
$\Rightarrow\ \frac{5\text{n}}{2}=18+47+\frac{5}{2}$
$\Rightarrow\ \frac{5\text{n}}{2}=\frac{36+94+5}{2}$
$\Rightarrow\ 5\text{n}=135$
$\Rightarrow\ \text{n}=\frac{135}{5}$
$\Rightarrow\ \text{n}=27$
Hence, Total $27$ terms in given A.P.
View full question & answer→Question 1223 Marks
The $n ^{\text {th }}$ term of an A.P. is $6 n +2$. Find the common difference.
AnswerIn the given problem, $n ^{\text {th }}$ term is given by " $a_n=6 n+2$ ". To find the common diffrence of the A.P., we need two consecutive terms of the A.P.
So, let us find the first and the second term of the given A.P.
First term $(n=1)$.
$a_1=6(1)+2$
$=6+2$
$=8$
Second term $(n=2)$,
$a_2=6(2)+2$
$=12+2$
$=14$
Now, the common difference of the A.P. (d) $=a_2-a_1$
$=14-8$
$=6$
Therefore, the common difference is $d =6$.
View full question & answer→Question 1233 Marks
If an A.P. consists of n terms with first term a and $n ^{\text {th }}$ terml show that the sum of the $m ^{\text {th }}$ term from the beginning and the $m ^{\text {th }}$ term from the end is $( a + l$ ).
AnswerIn the given problem, we have an A.P. which consists of n terms.
Here,
The first term (a) = a
The last term $\left(a_n\right)=1$
Now, as we know,
$a_n=a+(n-1) d$
So, for the $m ^{\text {th }}$ term from the beginning, we take $( n = m )$,
$a_m=a+(m-1) d$
$=a+m d-d \ldots . .(i)$
Similarly, for the $m ^{\text {th }}$ term from the end, we can take/ as the first term.
So, we get,
$a_m=1-(m-1) d$
$=1-m d+d \ldots . . . \text { (ii) }$
Now, we need to prove $a_m+a_m=a+1$
So, adding (i) and (ii), we get,
$a_m+a_m=(a+m d-d)+(l-m d+d)$
$=a+m d-d+1-m d+d$
$=a+1$
Therefore, $a_m+a_m=a+1$
Hence proved.
View full question & answer→Question 1243 Marks
Prove that no matter what the real numbers $a$ and $b$ are, the sequence with $n^{\text {th }}$ term $a+n b$ is always an A.P. What is the common difference?
AnswerGiven sequence $\left(a_n\right)=a_n+6 n$
$n ^{\text {th }}$ term $\left(a_n\right)=a+n b$
$(n+1)^{\text {th }}$ term $\left(a_{n+1}\right)=a+(n+1) b$.
Common difference $(d)=a_{n+1}-a_n$
$d=(a+(n+1) b)-(a+n b)$
$=a+n b+b-a-n b$
$=b$
$\therefore$ Common difference (d) does not depend on $n ^{\text {th }}$ value so, given sequence is in AP with $( d )= b$.
View full question & answer→Question 1253 Marks
Resham wanted to save at least 6500 for sending her daughter to school next year (after 12 months). She saved Rs. 450 in the first month and raised her savings by Rs. 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?
AnswerGiven,
Resham saved Rs. 450 in the first month and raised her saving by Rs. 20 every month and saved in next 12 months.
First term (a) = 450
Common difference (d) = 20
and No. of terms (n) = 12
We know sum of n terms is in A.P.
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}$
$\text{S}_\text{n}=\frac{12}{2}[2\times450+(12-1)\times20]$
$\Rightarrow\ \text{S}_\text{n}=6[900+220]$
$\Rightarrow\ \text{S}_\text{n}=6720$
Here we can see that Resham saved Rs. 6720 which is more than 6500.
So, yes Resham shall be able to send her daughter to school.
View full question & answer→Question 1263 Marks
Find the sum of the following arithmetic progressions:
$1, 3, 5, 7, .....$ to $12$ terms.
AnswerIn an A.P. let first term = a, common difference = d, and there are n terms.
Then, sum of n terms is,
$\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
Given,
$1, 3, 5, 7, .....$
Here,
First term $a = 1$
Difference $d = 3 - 1 = 2$
and no of terms $n = 12$
We know $\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
$\Rightarrow\ \text{S}_\text{n}=\frac{12}{2}[2(1)+(12-1)2]$
$\Rightarrow S_n = 6[2 + 11 \times 2]$
$\Rightarrow S_n = 6 \times 24$
$\Rightarrow S_n = 144$
Hence, sum of $12$ terms is $144.$
View full question & answer→Question 1273 Marks
Find the number of all three digit natural numbers which are divisible by $9$.
AnswerFirst three-digit number that is divisible by $9$ is $108.$
Next number is $108 + 9 = 117.$
And the last three-digit numbet that is divisible by $9$ is $999.$
Thus, the progression will be $108, 117, ....., 999.$
All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 180 and the common difference as 9.
We know that, $n^{th}$^ term $a_n = a + (n - 1)d$
According to the question,
$999 = 108 + (n - 1)9$
$\Rightarrow 108 + 9n - 9 = 999$
$\Rightarrow 99 + 9n = 999$
$\Rightarrow 9n = 999 - 99$
$\Rightarrow 9n = 900$
$\Rightarrow n = 100$
Thus, the number of all three digit natural numbers which are divisible by $9$ is $100.$
View full question & answer→Question 1283 Marks
Find the sum of two middle terms of the $\text{A.P.:}-\frac{4}{3},-1,\frac{-2}{3},-\frac{1}{3}, .....\ ,4\frac{1}{3}.$
Answer$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\text{a}=-\frac{4}{3},\text{d}=\frac{1}{3},\text{a}_\text{n}=\frac{13}{3}$
$\Rightarrow\ \text{a}+(\text{n}-1)\text{d}=\frac{13}{3}$
$\Rightarrow\ \Big(-\frac{4}{3}\Big)+(\text{n} - 1)\big(\frac{1}{3}\big)=\frac{13}{3}$
$\Rightarrow\ 13=-4+\text{n}-1$
$\Rightarrow\ \text{n}=18$
Midlle terms are $\frac{\text{n}}{2}^{\text{th}}\text{ and }\frac{\text{n}}{2}+1^{\text{th}},$ i.e., $9^{th}$ and $10^{th}$ terms.
$\text{a}_9=\text{a}+8\text{d}=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3}$
$\text{a}_{10}=\text{a}+9\text{d}=-\frac{4}{3}+\frac{9}{3}=\frac{5}{3}$
$\therefore\ \text{a}_9+\text{a}_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3$
View full question & answer→Question 1293 Marks
Find $n$ if the given value of $x$ is the $n ^{\text {th }}$ term of the given A.P.
$20,50,75,100, \ldots . . ; x=1000$.
AnswerWe have,
$25,50,75,100, \ldots ., x=1000$
First term a $=25$
Difference $d=50-25=25$
and Last term $a_n=1000$
We know
$a_n=a+(n-1) d$
$1000=25+(n-1) 25$
$1000=25+25 n-25$
$n=\frac{1000}{25}=40$
Hence, The value of $n$ is $40$ .
View full question & answer→Question 1303 Marks
The sum of first $m$ terms of an A.P. is $4 m^2-m$. If its $n^{\text {th }}$ term is 107 . find the value of $n$. Also, find the $21^{\text {st }}$ term of this A.P.
Answer$S_{m}=4 m^2-m, T_{n}=107$
$S_1=4(1)^2-1=4-1=3$
$S_2=4(2)^2-2=16-2=14$
$\therefore T_2=S_2-S_1=14-3=11$
$\text { and } a=S_1=3$
$d=t_2-t_1=11-3=8$
$\text { Now, } T_{n}=a+(n-1) d$
$107=3+(n-1) 8$
$\Rightarrow 107-3=(n-1) 8$
$\Rightarrow 104=(n-1) \times 8$
$n-1=\frac{104}{8}=13$
$\therefore n=13+1=14$
$T_{21}=a+(n-1) d$
$=3+(21-1) \times 8=3+20 \times 8$
$=3+160=163$
View full question & answer→Question 1313 Marks
The $19^{\text {th }}$ term of an A.P. is equal to three times its sixth term. If its $9^{\text {th }}$ term is 19 , find the A.P.?
AnswerLet a be the first term, d be the common difference, then
$T_{19}=3 . T_6$ and $T_9=19$
$T_n=a+(n-1) d$
Now, $T_{19}=a+(19-1) d=a+18 d$
$T_6=a+(6-1) d=a+5 d$
$T_9=19 \Rightarrow a+(9-1) d=19$
$\Rightarrow a+8 d=19 \ldots . . \text { (i) }$
$T_{19}=3 T_6$
$a+18 d=3(a+5 d)$
$a+18 d=3 a+15 d$
$18 d-15 d=3 a-a$
$3 d=2 a$
$2 a=3 d$
$\Rightarrow a=\frac{3}{2} d \ldots \text { (ii) }$
From (i), a $+8 d=19$
$\Rightarrow \frac{3}{2} d+8 d=19 \Rightarrow \frac{19}{2} d=19$
$\Rightarrow d=\frac{19 \times 2}{19}=2$
$\text { and } a=\frac{3}{2}, d=\frac{3}{2} \times 2=3$
$\therefore$ A.P. will be $3,5,7,9,11, \ldots .$.
View full question & answer→Question 1323 Marks
Show that the sum of all odd integers between $1$ and $1000$ which are divisible by $3$ is $83667$.
AnswerAll odd numbers divisible by $3$ between $1$ to
$1000$ will be $3, 9, 15, 21, ......, 999$
Where $a = 3, d = 9 - 3 = 6$ and $l = 999$
$\therefore$ $a_n = a + (n - 1)d$
$\Rightarrow 999 = 3 + (n - 1) \times 6$
$\Rightarrow 999 = 3 + 6n - 6 \Rightarrow 6n = 999 + 6 - 3$
$\Rightarrow\ 6\text{n}=1002\Rightarrow\ \text{n}=\frac{1002}{6}=167$
$\therefore$ Number of terms $= 167$
Now, $\text{S}_{167}=\frac{\text{n}}{2}[\text{a}+\text{l}]=\frac{167}{2}[3+999]$
$=\frac{167}{2}(1002)$
$=167(501)$
$=83667$
Hence proved.
View full question & answer→Question 1333 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$a + b, (a + 1) + b, (a + 1) + (b + 1), (a + 2) + (b + 1), (a + 2) + (b + 2), .....$
AnswerIn the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference (d),
$a + b, (a + 1) + b, (a + 1) + (b + 1), (a + 2) + (b + 1), (a + 2) + (b + 2), .....$
Let $a_1 = a + b$
$a_2 = (a + 1) + b$
$a_3 = (a + 1) + (b + 1)$
$a_4 = (a + 2) + (b + 1)$
$a_5 = (a + 2) + (b + 2)$
Now $a_2 - a_1 = (a + 1) + b - a - b$
$= a + 1 + b - a - b = 1$
$a_3 - a_2 = (a + 1) + (b + 1) - {(a + 1) + b}$
$= a + 1 + b + 1 - a - 1 - b = 1$
$a_4 - a_3 = {(a + 2) + (b + 1)} - {(a + 1) + (b + 1)}$
$= a + 2 + b + 1 - a - 1 - b - 1 = 1$
$a_5 - a_4 = {(a + 2) + (b + 2)} - {(a + 2) + (b + 1)}$
$= a + 2 + b + 2 - a - 2 - b - 1 = 1$
We see that common diffrence is $1$
$\therefore$ It is an A.P.
View full question & answer→Question 1343 Marks
Find the sum of the first 15 terms of each of the following sequences having $n^{th}$ term as
$a_n = 3 + 4n$
Answer$a_n = 3 + 4n,$ number of terms $= 15$
$a_1 = 3 + 4 \times 1 = 3 + 4 = 7$ or $a = 7$
$a_2 = 3 + 4 \times 2 = 3 + 8 = 11$
$\therefore$ $d = a_2 - a_1 = 11 - 7 = 4$
Now $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{15}=\frac{15}{2}[2\text{a}+(15-1)\text{d}]$
$=\frac{15}{2}[2\times7+(15-1)\times4]$
$=\frac{15}{2}[14+14\times4]=\frac{15}{2}[14+56]$
$=\frac{15}{2}\times70=15\times35=525$
View full question & answer→Question 1353 Marks
The sum of three numbers in A.P. is $12$ and the sum of their cubes is $288$. Find the numbers.
AnswerLet the three numbers in A.P. be
$a - d, a, a + d$
$\therefore$ $a - d + a + a + d = 12 \Rightarrow 3a = 12$
$\Rightarrow\ \text{a}=\frac{12}{3}=4$
and $(a - d)^3 + a^3 + (a + d)^3 = 288$
$a^3 - 3a^2d + 3ad^2 - d^3 + a^3 + a^3 + 3a^2d + 3ad^2 + d^3 = 288$
$3a^3 + 6ad^2 = 288 \Rightarrow 3(4)^3 + 6 \times 4d^2 = 288$
$192 + 24d^2 = 288 \Rightarrow 24d^2 = 288 - 192 = 96$
$\Rightarrow\ \text{d}^2=\frac{96}{24}=4=(\pm2)^2$
$\therefore\ \text{d}=\pm2$
$\therefore$ Number will be $4 - 2, 4, 4 + 2$ or $2, 4, 6$ or $4 + 2, 4, 4 - 2$ or $6, 4, 2$.
View full question & answer→Question 1363 Marks
Write the sequence with $n^{th}$ term:
$a_n = 3 + 4n.$
Show the all of the above sequences form A.P.
Answer$a_n = 3 + 4n$
Put $n = 1, 2, 3...$
$a_1 = 3 + 4 \times 1 = 7$
$a_2 = 3 + 4 \times 2 = 11$
$a_3 = 3 + 4 \times 3 = 15$
Now, $a_2 - a_1 = a_3 - a_2$
$11 - 7 = 15 - 11$
$4 = 4$
Common difference is $4$
So that A.P
$7, 11, 15, 19...$
View full question & answer→Question 1373 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$a_n = 2n^2 - 3n + 1.$
Answer$a_n = 2n^2 - 3n + 1$.The given sequence is $a_n = 2n^2 - 3n + 1$.
To write first tive terms of given sequence an, we put $n = 1, 2, 3, 4, 5$. Then we get
$a_1 = 2.1^2 - 3.1 + 1 = 2 - 3 + 1 = 0$
$a_2 = 2.2^2 - 3.2 + 1 = 8 - 6 + 1 = 3$
$a_3 = 2.3^2 - 3.3 + 1 = 18 - 9 + 1 = 10$
$a_4 = 2.4^2 - 3.4 + 1 = 32 - 12 + 1 = 21$
$a_5 = 2.5^2 - 3.5 + 1 = 50 - 15 + 1 = 36$
$\therefore$ The required first five tms of given sequence $a_n = 2n^2 - 3n + 1$ are $0, 3, 10, 21, 36.$
View full question & answer→Question 1383 Marks
A thief, after committing a theft runs at a uniform speed of $50m/ minute$. After $2$ minutes, a policeman runs to catch him. He goes $60$ min first minute and increases his speed by $5m/ minute$ every succeeding minute. After how many minutes, the policeman will catch the thief?
AnswerLet total time be $22$ minutes.
Total distance covered by thief in $22$ minutes = Speed $\times $ Time
$= 100 \times n = 100n\ metres$
Total distance covered by policeman
$1^{\text{st}}\text{ min.}+2^{\text{nd}}\text{ min.}+3^\text{rd}\text{ min}+\ .....(\text{n}-1)\text{terms} \\ 100 \ \ \ \ \ \ \ \ \ \ \ \ 110 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 120 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
$\begin{bmatrix}\because\ \text{Thirf runs = n miss}\ \ \ \ \ \ \ \ \ \ \ \ \\ \text{Policeman runs = (n - 1)mins} \end{bmatrix}$
Here,
$a = 100, d = 110 - 100 = 10, 'n' = n - 1$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\therefore\ 100\text{n}=\frac{(\text{n}-1)}{2}[2(100)+(\text{n}-1-1)(10)]$
$\Rightarrow (n - 1) [200 + 10n - 20] = 200n$
$\Rightarrow (n - 1) [10n + 180] = 200n$
$\Rightarrow 10n^2 + 180n - 10n - 180 - 200n = 0$
$\Rightarrow 10n^2 - 30n - 180 = 0$
$n^2 - 3n = 18 = 0$ [Dividing both sides by $10$]
$\Rightarrow n^2 - 6n + 3n - 18 = 0$
$\Rightarrow n(n - 6) + 3(n - 6) = 0$
$\Rightarrow (n + 3)(n - 6) = 0$
$\Rightarrow n + 3 = 0$ or$ n - 6 = 0$
$\Rightarrow n = -3$(reject) or $n = 6$
Since n (time) cannot be negative.
$\therefore$ Time taken by policeman to cathc the thief
$= n - 1 = 6 - 1 = 5$ minutes.
View full question & answer→Question 1393 Marks
Find the sum of first n odd natural numbers.
AnswerIn this problem, we need to find the sum of first n odd natural numbers.
So, we know that the first odd natural number is $1$. Also, all the odd terms will form an A.P. with the common difference of $2$.
So here,
First term $(a) = 1$
Common difference $(d) = 2$
So, let us take the number of terms as n
Now, as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
So, for n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2(1)+(\text{n}-1)2]$
$=\frac{\text{n}}{2}[2+2\text{n}-2]$
$=\frac{\text{n}}{2}(2\text{n})$
$=\text{n}^2$
Therefore, the sum of first n odd natural numbers is $S_n = n^2$.
View full question & answer→Question 1403 Marks
Find n if the given value of $x$ is the $n^{th}$ term of the given A.P.
$1,\frac{21}{11},\frac{31}{11},\frac{41}{11}, .....;\text{ x}=\frac{171}{11}.$
AnswerGiven A.P.
$1,\frac{21}{11},\frac{31}{11},\frac{41}{11}, .....\frac{171}{11}$
First term $a = 1$
Difference $\text{d}=\frac{21}{11}-1=\frac{21-11}{11}=\frac{10}{11}$
and last term $\text{a}_\text{n}=\frac{171}{11}$
We know
$a_n =a + (n - 1)d$
$\Rightarrow\ \frac{171}{11}=1+(\text{n}-1)\frac{10}{11}$
$\Rightarrow\ \frac{171}{11}=1+\frac{10}{11}\text{n}-\frac{10}{11}$
$\Rightarrow\ \frac{10}{11}\text{n}=\frac{171}{11}-1+\frac{10}{11}$
$\Rightarrow\ \frac{10\text{n}}{11}=\frac{171-11+10}{11}\Rightarrow\ 10\text{n}=170$
$\Rightarrow\ \text{n}=17$
Hence, value of $n$ is $17$.
View full question & answer→Question 1413 Marks
The first term of an A.P. is p and its common difference is q. Find its $10^{th}$ term.
AnswerGiven,
First term, $a = p$
and Difference, $d = q$
We have to find $10^{th}$ term,
We knaw $a_n = a + (n - 1)d$
$\Rightarrow a_{10} = p + (10 - 1)q$
$\Rightarrow a_{10} = p + 9q$
Hence, term of given A.P. is $p + 9q.$
View full question & answer→Question 1423 Marks
The $7^{\text {th }}$ term of an A.P. is $32$ and its $13^{\text {th }}$ term is $62$ . Find the A.P.
AnswerGiven, $a=32 a+(7-1) d=32 a+6 d=32 \ldots$....(i) and
$a_{13}=62 a+(13-1) d=62 a+12 d=62$
Subtract (1) from (2)
$d =\frac{30}{6}=5$ Put $d =5$ in $a +6 d=32 a +6 \times 5=32 a =2$
Then the sequence is $a, a+d, a+2 d, a+3 d, \ldots . .$
$\Rightarrow 2,7,12,17, \ldots .$. View full question & answer→Question 1433 Marks
Write the value of $x$ for which $2x, x + 10$ and $3x + 2$ are in A.P.
AnswerHere, we are given three terms, First term $\left(a_1\right)=2 x$
Second term $\left(a_2\right)=x+10$ Third term $\left(a_3\right)=3 x+2$
We need to find the value of $x$ for which these terms are in A.P.
So, in an A.P. the difference of two adjacent terms is always constant.
So, we get, $d=a_2-a_1 d=(x+10)-(2 x) d=x+10-2 x d=10-x$.....(i)
Also, $d=a 3-a 2 d=(3 x+2)-(x+10)$
$d=2 x-8 \ldots . . . \text { (ii) }$
Now, on equatin (i) and (ii),
we get, $10-x=2 x-82 x+x=10+83 x=18$
$x=\frac{18}{3} x=6$ Therefore, for $x=6$, these three terms will from an A.P.
View full question & answer→Question 1443 Marks
Find:
Is $68$ a term of the A.P. $7, 10, 13, ...?$
AnswerIn the given problem, we are given an A.P. and the Value of one of its term.
We need to find whether it is a term of the A.P. or not so here we will use the formula $a_n = a + (n - 1)d.$
Here,
A.P is $7, 10, 13, .....$
$a_n = 68, a = 7$ and $d = 10 - 7 = 3$
Using the above mentioned formula, we get
$68 = 7 + (n - 1)3$
$\Rightarrow 68 - 7 = 3n - 3$
$\Rightarrow 61 + 3 = 3n$
$\Rightarrow 64 = 3n$
$\Rightarrow\ \text{n}=\frac{64}{3}$
Since, the value of n is a fraction.
Thus, 68 is not team of the given A.P.
View full question & answer→Question 1453 Marks
Find the sum of the following arithmetic progressions:
-26, -24, -22, ..... to 36 terms.
AnswerIn an A.P. let first term = a, common difference = d, and there are n terms. Then, sum of n terms is,
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a} + (\text{n} - 1)\text{d}\big]$
Given expression -26, -24, -22, ..... to 36 terms
First term (a) = -26
Common difference (d) = -24 - (-26) = -24 + 26 = 2
Sum of n terms $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a} + (\text{n} - 1)\text{d}\big]$
Sum of n terms $\text{S}_{36}=\frac{36}{2}\big[2(-26)+(36-1)2\big]$
= 18[-52 + 70]
= 18 × 18
= 324
$\therefore\ \text{S}_\text{n}=324$
View full question & answer→Question 1463 Marks
Find the sum of all natural numbers between $1$ and $100$, which are divisible by $3$.
AnswerNatural number which are divisible by $3$
Between $1$ to $100$ are $3, 6, 9, ....., 96, 99$
Whose first term $(a) = 3$
and common difference $(d) = 6 - 3 = 3$
$a_n = a + (n - 1)d$
$99 = 3 + (n - 1) \times 3$
$99 = 3 + 3n - 3 \Rightarrow 99 = 3n$
$\therefore\ \text{n}=\frac{99}{3}=33$
Now $\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{33}{2}[3+99]=\frac{33}{2}\times102$
$=33\times51=1683$
View full question & answer→Question 1473 Marks
Find the sum of the first $25$ terms of an A.P. whose $n^{\text {th }}$ term is given by $a_n=7-3 n$.
AnswerHere, we are given an A.P. whose $n^{\text {th }}$ term given by the following expression, $a_n=7-3 n$. We need to fint the sum of first $25$ terms.
So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_n=\left(\frac{n}{2}\right)(a+1)$
Where, $a =$ the first term
I = the last term
So, for the given A.P.
the first term (a) will be calculated using $n=1$ in the given equation for $n$th term of A.P.
$a=7-3(1)$
$=7-3$
$=4$
Now, the last term ( 1 ) of the nth term is given
$1=a_n=7-3 n$
So, on substituting the values in the formula for the sum of $n$ term of an A.P., we get
$S_{25}=\left(\frac{25}{2}\right)[(4)+7-3(25)]$
$=\left(\frac{25}{2}\right)[11-75]$
$=\left(\frac{25}{2}\right)(-64)$
$=(25)(-32)$
$=-800$
Therefore, the sum of the $25$ terms of the given A.P. is $S _n=-800$.
View full question & answer→Question 1483 Marks
$51$ terms of the A.P. whose second term is $2$ and fourth term is $8$.
AnswerGiven, $a_2 = 2$ and
$a_4 = 8 a + d = 2 .....(i) $
$a + 3d = 8 .....(ii)$
Put $d = 3 in .....(i) \Rightarrow a + d = 2 a + 3 = 2 a = -1$
$\text{S}_{51}=\frac{51}{2}(2\times-1+(51-1)\times3)$
$\Big(\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})\Big)$
$=\frac{51}{2}(-2+50\times3)$
$=\frac{51}{2}\times148$ $=3774$
$\therefore\ \text{S}_{\text{n}}=3774$ View full question & answer→