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 $n^{\text {th }}$ term of an A.P is given by $(-4 n+15)$, 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 $S_n = 636$
$636=\frac{\text{n}}{2}(2.9+(\text{n}-1)8)$
$1272 = n(18 - 8 + 8n)$
$1272 = n(10 + 8n)$
$2 \times 636 = 2n(5 + 4n)$
$636 = 5n + 4n^2$
$4n^2 + 5n - 636 = 0$
$4n^2 + 53n - 48n - 636 = 0$
$4n^2 - 48n + 53n - 636 = 0$
$4n(n - 12) + 53(n - 12) = 0$
$(4n + 53)(n - 12) = 0$
$\therefore\ \text{n}=12$ (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 $n^{\text {th }}$ term of the given A.P.
$-1,-3,-5,-7, \ldots \ldots ; x=-151$
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. $3, 15, 27, 39, ...$ will be $120$ more than its $21^{st} $term?
AnswerGiven,
A.P. is $3, 15, 27, 39, .....$
Let $n^{th}$ term is 120 more than $21^{st}$ term
Then $a_n = 120 + a_{21}$
For the given sequence
$a = 3, d = 15 - 3 = 12$
$a + (n - 1)d = 120 + a + (21 - 1)d$
$(n - 1)12 = 120 + 20(12)$
$(n - 1)12 = 360$
$(\text{n}-1)=\frac{360}{12}=30$
$n = 31$
$\therefore 31^{\text {st }}$ 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 $3, 8, 13, .....$
$a_n = 248$
$a = 3$
Now,
Common difference $(d) = a_1 - a$
$= 8 - 3$
$= 5$
Thus, using the above mentioned formula
$a_n = a + (n - 1)d$
$248 = 3 + (n - 1)5$
$248 - 3 = 5n - 5$
$245 + 5 = 5n$
$\text{n}=\frac{250}{5}$
$n = 50$
Thus, $n = 50$
Therefore 248 is the $50^{th}$ 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 $S_1+S_3=2 S_2$.
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 $5$.
AnswerSince, multiples of 2 as well as of $5=\operatorname{LCM}$ of $(2,5)=10$
$\therefore$ 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
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]\ .....\text{(i)}$
$\because$ $a_n = a + (n - 1)d = l$
$\Rightarrow 10 + (n - 1)10 = 490$
$\Rightarrow (n - 1)10 = 480$
$\Rightarrow n - 1 = 48$
$\Rightarrow n = 49$
From Eq. (i), $\text{S}_{49}=\frac{49}{2}(10+490)$
$=\frac{49}{2}\times500$
$=49\times250=12250$
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 $3 n^2+4 n$. Find the $25^{\text {th }}$ term of this A.P.
Answer$S_n=3 n^2+4 n$
We know
$a_n=S_n-S_{n-1}$
$\therefore a_n=3 n^2+4 n-3(n-1)^2-4(n-1)$
$\Rightarrow a_n=6 n+1$
$a_{25}=6(25)+1=151$
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,
$a_n = (3n^2 + 5n) - [3(n - 1)^2 + 5(n - 1)]$
$= 3n^2 + 5n - [3(n^2 + 1 - 2n) + 5n - 5] [Using(a - b)^2 = a^2 + b^2 - 2ab]$
$= 3n^2 + 5n - (3n^2 + 3 - 6n + 5n - 5)$
$= 3n^2 + 5n - 3n^2 - 3 + 6n - 5n + 5$
$= 6n + 2 .....(i)$
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 $5^{\text {th }}$ term of an $A.P$. is $31$ and $25^{\text {th }}$ term is $140$ more than the $5^{\text {th }}$ 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$S_q = 63q - 3q^2$
We know
$a_q = S_q - S_{q-1}$
$\therefore$ $a_q = 63q - 3q^2 - 63(q - 1) + 3(q - 1)^2$
$a_q = 66 - 6q$
Now, $a_p = -60$
$\Rightarrow 60 - 6p = -60$
$\Rightarrow 126 = 6p$
$\Rightarrow p = 21$
$a_{11} = 66 - 6 \times 11 = 0$
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 $n^{th}$ term of an A.P. is $6n + 2$. 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 $n ^{\text {th }}$ term/ 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 $(a_n) = l$
Now, as we know,
$a_n = a + (n - 1)d$
So, for the $m^{th}$ term from the beginning, we take (n = m),
$a_m = a + (m - 1)d$
$= a + md - d .....(i)$
Similarly, for the $m^{th}$ term from the end, we can take/ as the first term.
So, we get,
$a_{m'} = l - (m - 1)d$
$= l - md + d ..... (ii)$
Now, we need to prove $a_m + a_{m'} = a + l$
So, adding (i) and (ii), we get,
$a_m + a_{m'} = (a + md - d) + (l - md + d)$
$= a + md - d + l - md + d$
$= a + l$
Therefore, $a_m + a_{m'} = a + l$
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 $a + nb$ is always an A.P. What is the common difference?
AnswerGiven sequence $\left(a_n\right)=a_n+6 n$
$n^{\text {th }} \text { term }\left(a_n\right)=a+n b$
$(n+1)^{\text {th }} \text { term }\left(a_{n+1}\right)=a+(n+1) b$
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.
AnswerSOLution First 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 $108,117, \ldots . . ., 999$.
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, $n^{\text {th }}$ term $a_n=a+(n-1) d$
According to the question,
$999=108+(n-1) 9$
$\Rightarrow 108+9 n-9=999$
$\Rightarrow 99+9 n=999$
$\Rightarrow 9 n=999-99$
$\Rightarrow 9 n=900$
$\Rightarrow n=100$
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 $n^{th}$ term of the given $A.P.$
$20, 50, 75, 100, .....; x = 1000.$
AnswerWe have,
$25, 50, 75, 100, ...., x = 1000$
First term $a = 25$
Difference $d = 50 - 25 = 25$
and Last term $a_n = 1000$
We know
$a_n = a + (n - 1)d$
$1000 = 25 + (n - 1)25$
$1000 = 25 + 25n - 25$
$\text{n}=\frac{1000}{25}=40$
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$S_m = 4m^2 - m, T_n = 107$
$S_1 = 4(1)^2 - 1 = 4 - 1 = 3$
$S_2 = 4(2)^2 - 2 = 16 - 2 = 14$
$\therefore$ $T_2 = S_2 - S_1 = 14 - 3 = 11$
$and a = S_1 = 3$
$d = t_2 - t_1 = 11 - 3 = 8$
Now, $T_n = a + (n - 1)d$
$107 = 3 + (n - 1)8$
$\Rightarrow 107 - 3 = (n - 1)8$
$\Rightarrow 104 = (n - 1) \times 8$
$\text{n}-1=\frac{104}{8}=13$
$\therefore$ $n = 13 + 1 = 14$
$T_{21} = a + (n - 1)d$
$= 3 + (21 - 1) \times 8 = 3 + 20 \times 8$
$= 3 + 160 = 163$
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 1$9$ , 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, $T_{19} = a + (19 - 1)d = a + 18d$
$T_6 = a + (6 - 1)d = a + 5d$
$T_9 = 19 \Rightarrow a + (9 - 1)d = 19$
$\Rightarrow a + 8d = 19 .....(i)$
$T_{19} = 3T_6$
$a + 18d = 3(a + 5d)$
$a + 18d = 3a + 15d$
$18d - 15d = 3a - a$
$3d = 2a$
$2a = 3d$
$\Rightarrow\ \text{a}=\frac{3}{2}\text{d}\ ...(\text{ii})$
From $(i), a + 8d = 19$
$\Rightarrow\ \frac{3}{2}\text{d}+8\text{d}=19\Rightarrow\ \frac{19}{2}\text{d}=19$
$\Rightarrow\ \text{d}=\frac{19\times2}{19}=2$
and $\text{a}=\frac{3}{2},\text{d}=\frac{3}{2}\times2=3$
$\therefore$ $A.P$. will be $3, 5, 7, 9, 11, .....$
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 × Time
$= 100 \times n = 100$n 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, $10^{th}$ 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 $
$= 32 a + 6d = 32 .....(i)$ and
$a_{13} = 62 a + (13 - 1)d$
$= 62 a + 12d = 62 .....(ii)$ Subtract (1) from (2)
$\text{d}=\frac{30}{6}=5$
Put $d = 5$ in $a + 6d = 32 a + 6 × 5 = 32 a = 2$
Then the sequence is $a, a + d, a + 2d, a + 3d, .....$
$\Rightarrow 2, 7, 12, 17, .....$ 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 ($a_1$)
$= 2x$ Second term $(a_2)$
$= x + 10$ Third term $(a_3)$
$= 3x + 2$
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, $d = a_2 - a_1 d$
$= (x + 10) - (2x) d$
$= x + 10 - 2x d$
$= 10 - x .....(i)$
Also, $d = a3 - a2d$
$= (3x + 2) - (x + 10)$
$d = 2x - 8 .....(ii)$
Now, on equatin (i) and (ii),
we get, $10 - x = 2x - 8 2x + x $
$= 10 + 8 3x = 18$
$\text{x}=\frac{18}{3}$ 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, $\text{S}_\text{n}=\Big(\frac{\text{n}}{2}\Big)(\text{a}+\text{l})$
Where, a = the first term
l = the last term
So, for the given $A.P$.
the first term (a) will be calculated using $n = 1$ in the given equation for nth term of $A.P$.
$a = 7 - 3(1)$
$= 7 - 3$
$= 4$
Now, the last term (l) of the nth term is given
$l = a_n = 7 - 3n$
So, on substituting the values in the formula for the sum of n term of an $A.P$., we get
$\text{S}_{25}=\Big(\frac{25}{2}\Big)[(4)+7-3(25)]$
$=\Big(\frac{25}{2}\Big)[11-75]$
$=\Big(\frac{25}{2}\Big)(-64)$
$=(25)(-32)$
$=-800$
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→