Question 12 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$3, 3, 3, 3, .....$
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,
$3, 3, 3, 3, .....$
Here, $a_1 = 3, a_2 = 3, a_3 = 3, a_4 = 3$
Difference between terms,
$d_1 = a_2 - a_1 = 3 - 3 = 0$
$d_2 = a_3 - a_2 = 3 - 3 = 0$
and $d_3 = a_4 - a_3 = 3 - 3 = 0$
Hence, common difference is 0 then we say sequence in A.P.
View full question & answer→Question 22 Marks
Find the sum of the first $25$ terms of an $A.P$. whose $n^{th}$ term is given by $a_n = 2 - 3n.$
AnswerGiven,
$n^{th}$ term $a_n = 2 - 3n$
Put $n = 1, a_1 = 2 - 3 \times 1 = -1$
Put $n = 25, a_{15} = l = 2 - 3 \times 15 = -43$
$\therefore \text{S}_{25}=\frac{25}{2}(-1-43)=\frac{25}{2}(-44)=-925$
$\therefore\ \text{S}_{25}=-925$
View full question & answer→Question 32 Marks
Find the sum:
$18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$
Answer$\text{S}=18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$Here, a = 18, $\text{l}=\text{a}_\text{n}=-49\frac{1}{2}=\frac{-99}{2}$
$\text{d}=\frac{31}{2}-18=\frac{-5}{2}$ For n $\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$ $\frac{-99}{2}=18+(\text{n}-1)\Big(\frac{-5}{2}\Big)$ $\frac{-99}{2}=\frac{36-(\text{n}-1)5}{2}$ $-99=36-(\text{n}-1)5$ $(\text{n}-1)5=36+99$ $(\text{n}-1)5=135$ $\text{n}-1=\frac{135}{5}$ $\text{n}-1=27$ $\text{n}=27+1$ $\text{n}=28$ Now, $\text{S}_{28}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $=\frac{28}{2}\Big[2\times18+(28-1)\times\Big(\frac{-5}{2}\Big)\Big]$ $=14\Big[36+27\times\Big(\frac{-5}{2}\Big)\Big]$ $=14\Big[36-\frac{135}{2}\Big]$ $=14\Big[\frac{72-135}{2}\Big]=7\times(-63)=-441$
View full question & answer→Question 42 Marks
The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n.
AnswerAccording to the question, we have
$\frac{\text{n}}{2}[2(8)+(\text{n}-1)20]=\frac{2\text{n}}{2}[2(-30)+(2\text{n}-1)8]$
⇒ [16 + 20n - 20] = 2[-60 + 16n - 8]
⇒ 20n - 4 = -136 + 32n
⇒ 12n = 132
⇒ n = 11.
View full question & answer→Question 52 Marks
Find the $8^{th}$ term from the end of the $A.P. 7, 10, 13,... 184.$
AnswerThe given A.P. is $7, 10, 13 ...... 184$
Here $a = 7$
Common difference $d = 10 - 7 = 3$
ast term $l = 184$
Now, $n^{th} $ term
$a_n = a + (n - 1)d$
$184 = 7 + (n - 1)3$
$184 - 7 = (n - 1)3$
$177 = (n - 1)3$
$\text{n}-1=\frac{177}{3}$
$n - 1 = 59$
$n = 59 + 1$
$n = 60$
Now $8^{th} $term end is $53$
$a_{53} = a + (53 - 1)d$
$a_{53} = 7 + 52 \times 3$
$a_{53} = 7 + 156$
$a_{53} = 163$
View full question & answer→Question 62 Marks
Find the sum of last ten terms of the A.P. 8, 10, 12, 14,…, 126.
AnswerThe given A.P. is 8, 10, 12, 14, ....., 126.
a = 8 and d = 2.
When this A.P. is reversed, we get the A.P.
126, 124, 122, 120, .....
So, first term becomes 126 and common difference -2.
The sum of first 10 terms of this A.P. is as follows,
$\text{S}_{10}=\frac{10}{2}[2\times126+9(-2)]$
$=5[234]$
$=1170$
View full question & answer→Question 72 Marks
How many terms are there in the A.P.?
$7, 10, 13, ..... 43.$
Answer$7, 10, 13, ..... ,73.$
From given $A.P.$
$a = 7, d = 10 - 7 = 3, a_n = a + (n - 1)d.$
Let a_n = 43 (last term)
$7 + (n - 1)3 = 43$
$(\text{n}-1)=\frac{36}{3}=12$
$n = 13$
$\therefore$ 13 terms are there in given A.P.
View full question & answer→Question 82 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.k, if $S_n=3 n^2+5 n$ and $a k=164$.
Answer$a_k=S_k-S_{k-1}$
$\Rightarrow 164=\left(3 k^2+5 k\right)-\left(3(k-1)^2+5(k-1)\right)$
$\Rightarrow 164=3 k^2+5 k-3 k^2+6 k-3-5 k+5$
$\Rightarrow 164=6 k+2$
$\Rightarrow 6 k=162$
$\Rightarrow k=27$
View full question & answer→Question 92 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$a_n = 3^n.$
Answer$a_n = 3^n$
Given sequence $a_n = 3^n$
To write first five terms of given sequence, put $n = 1, 2, 3, 4, 5$ in given sequence.
Then,
$a_1 = 3^1 = 3;$
$a_2 = 3^2 = 9;$
$a_3 = 3^3 = 27;$
$a_4 = 3^4 = 81;$
$a_5 = 3^5 = 243.$
View full question & answer→Question 102 Marks
Find:
$8^{th}$term of the$ A.P. 117, 104, 91, 78, ......$
AnswerGiven,
$A.P. 117, 104, 91, 78, ......$
Here,
First term $(a) = 117$
Common difference of the A.P. $(d) = 104 - 117$
$= -13$
Now, as we know,
$a_n = a + (n - 1)d$
So, for $8^{th}$^ term.
$a_8 = a + (8 - 1)d$
$= 117 + (7)(-13)$
$= 117 - 91$
$= 26$
Therefore, the $8^{th}$ term of the given A.P. is $a_8 = 26.$
View full question & answer→Question 112 Marks
For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k - 1 form on A.P.?
Answer(3k + 3) - (2k + 1) = (5k - 1) - (3k + 3)
3k + 3 - 2k - 1 = 5k - 1 - 3k - 3
k + 2 = 2k - 4
2k - k = 2 + 4
k = 6
View full question & answer→Question 122 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$10, 10 + 2^5, 10 + 2^6, 10 + 2^7, .....$
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),
$10, 10 + 2^5, 10 + 2^6, 10 + 2^7, .....$
In the given sequence
$a_1 = 10, a_2 = 10 + 2^5, a_3 = 10 + 2^6, a_4 = 10 + 2^7$
Check the conditon
$a_2 - a_1 = a_3 - a_2$
$10 + 2^5 - 10 = 10 + 2^6 - 10 - 2^5$
$2^5\neq2^6-2^5.$
$\therefore$ The given sequence is not in A.P.
View full question & answer→Question 132 Marks
Write the $n ^{\text {th }}$ term of an A.P. the sum of whose $n$ terms is $S_n$.
AnswerSum of $n$ terms $= S _{ n }$
Let a be the first term and $d$ be the common difference $a_n=S_n-S_{n-1}$.
View full question & answer→Question 142 Marks
Find the sum:
$3 + 11 + 19 + ..... + 803.$
Answer$3 + 11 + 19 + ..... + 803$
$a = 3, d = 11 - 3 = 8, l = a_n = 803$
$803 = 3 + (n - 1)8$
$\frac{800}{8}=\text{n}-1$
$\text{n}=101$
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\text{S}_{101}=\frac{101}{2}(3+803)$
$\text{S}_{101}=\frac{101}{2}(806)$
$\text{S}_{101}=101\times403$
$=40703$
View full question & answer→Question 152 Marks
Write the arithmetic progression when first term a and common difference d are as follows:
a = 4, d = -3.
AnswerFirst term (a) = 4
and common difference (d) = -3
Second term = a + d = 4 - 3 = 1
Third term = a + 2d = 4 + 2(-3) = 4 - 6 = -2
Fourth term = a + 3d = 4 + 3(-3) = 4 - 9 = -5
Fifth term = a + 4d = 4 + 4(-3) = 4 - 12 = -8
AP will be 4, 1, -2, -5, -8, ....
View full question & answer→Question 162 Marks
Find:
$10^{th}$ term of the $A.P. 1, 4, 7, 10, .....$
AnswerGiven A.P. is
$1, 4, 7, 10, ........$
First term $(a) = 1$
Common difference (d) = second term first term
$= 4 - 1$
$= 3.$
$n^{th}$ term in an $A.P. = a + (n - a)d$
$10^{th}$ term in an $1 + (10 - 1)3$
$= 1 + 9 × 3$
$= 1 + 27$
$= 28$
View full question & answer→Question 172 Marks
Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.
AnswerSuppose three parts of 207 are (a - d), a, (a + d) such that, (a + d) > a > (a - d).
a - d + a + a + d = 207
⇒ 3a = 207
⇒ a = 69
Now, (a - d) × a = 4623
⇒ 69(69 - d) = 4623
⇒ (69 - d) $= \frac{4623}{69}=67$
⇒ (69 - d) = 67
⇒ d = 2
Therefore, the three required parts are 67, 69 and 71.
View full question & answer→Question 182 Marks
Define an arithmetic progression.
AnswerA sequence $a_1, a_2, a_3, \ldots .$, an is called an arithmetic progression of then exists a constant $d$ Such that $a_2-a_1=d, a_3-a_2=d, \ldots . a_n-a_{n-1}=d$ and so on and d is called common difference.
View full question & answer→Question 192 Marks
Find the $12^{\text {th }}$ term from the end of the A.P. $-2,-4,-6, \ldots,-100$.
AnswerConsider the $A.P. -2, -4, -6, ....., -100.$
$a = -2, d = -2, a_n = -100$
$\Rightarrow a + (n - 1)d = -100$
$\Rightarrow (-2) + (n - 1)(-2) = -100$
$\Rightarrow 2n = 100$
$\Rightarrow n = 50$
The $12^{\text {th }}$ term the end will be the $39^{\text {th }}$ term from the starting.
$\therefore a_{39}=a+38 d=-2+38(-2)=-78 .$
View full question & answer→Question 202 Marks
Find the $12^{th}$ term from the end of the following arithmetic progressions:
$3, 5, 7, 9, ....., 201.$
Answer$3,5,7,9, \ldots . ., 2 d$
First term (a) = 3
Common difference ( d ) $=5-3=2$
$12^{\text {th }}$ term from the end is can be considered as (1) last term $=$ first term and common difference $=d^1=-d n^{\text {th }}$ term from the end $=$ last term $+(n-1)-d$,
$12^{\text {th }}$ term from end $=201+(12-1)(-2)$
$= 201 - 22$
$= 179$
View full question & answer→Question 212 Marks
Sum of 11 terms of the$ A.P. 2, 6, 10, 14.$
AnswerGiven,
$A.P., 2, 6, 10, 14, .....$
$a = 2, d = 4, S_n = S_{11}$
$=\frac{11}{2}(2.2+(11-1)4)$
$\Big(\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})\Big)$
$\frac{11}{2}(4+40)$
$=\frac{11}{2}\times44$
$\therefore\ \text{S}_{11}=242$
View full question & answer→Question 222 Marks
Justify whether it is true to say that the sequence, having following $n^{th}$ term is an$ A.P.$
$a_n = 1 + n + n^2.$
AnswerConsider the expression an $= 1 + n + n^2,$
For $n = 1, a_1 = 1 + 1 + 1 = 3$
For $n = 2, a_2 = 1 + 2 + 4 = 7$
For $n = 3, a_3 = 1 + 3 + 9 = 13$
For $n = 4, a_4 = 1 + 4 + 16 = 21$
The first four terms are $3, 7, 13, 21.$
The difference between each consecutive term in not same.
Hence this is not an A.P.
View full question & answer→Question 232 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$-225, -425, -625, -825, .....$
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),
$-225, -425, -625, -825, .....$
Here, $a_1 = -225, a_2 = -425, a_3 = -625, a_4 = -825$
Difference between terms,
$d_1= a_2 - a_1 = -425 - (-225) = -425 + 225 = -200,$
$d_2 = a_3 - a_2 = -625 - (-425) = -625 + 425 = -200,$
and $d_3 = a_4 - a_3 = -825 - (-625) = -825 + 625 = -200$
Hence, common difference is $-200$ then we say sequance in an $A.P.$
View full question & answer→Question 242 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$0, -4, -8, -12, .....$
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, $0, -4, -8, -12, .....$
Here, $a_1 = 0, a_2 = -4, a_3 = -8, a_4 = -12$
Difference between terms, $d_1 = a_2 - a_1 = -4 - 0 = -4,$
$d_2 = a_3 - a_2 = -8 - (-4) = -8 + 4 = -4, and d_3 = a_4 - a_3 = -12 - (-8) = -12 + 8 = -4$
Hence, common difference is -4 then we say sequance in an A.P.
View full question & answer→Question 252 Marks
Find:
$18^{th}$ term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},\ .....$
Answer$\text{A.P.}=\sqrt{2},3\sqrt{2},5\sqrt{2},\ .....$
Here,
First term $\text{(a)}=\sqrt{2}$
and Common difference $=3\sqrt{2}-\sqrt{2}$
$=2\sqrt{2}\text{ or }5\sqrt{2}-3\sqrt{2}=2\sqrt{2}$
$\text{Now }\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\therefore\ \text{a}_{18}=\sqrt{2}+(18-1)\times2\sqrt{2}$
$=\sqrt{2}+17\times2\sqrt{2}$
$=\sqrt{2}+34\sqrt{2}=35\sqrt{2}$
View full question & answer→Question 262 Marks
Find the sum:
$34 + 32 + 30 + ..... + 10.$
Answer$34 + 32 + 30 + ..... + 10$
$a = 34, d = -2, 1 = a_n = 10$
$10 = 34 + (n - 1) (-2)$
$+24 = 2(n - 1)$
$n - 1 = 12$
$n = 13$
$\therefore\ \text{S}_{13}=\frac{13}{2}(34+10)$
$=\frac{13}{2}\times44$
$=286$
View full question & answer→Question 272 Marks
Find the sum:
$25 + 28 + 31 + ..... + 100.$
Answer$25 + 28 + 31 + ..... + 100$
$a = 25, d = 3, l = a_n = 100$
$100 = 25 + (n - 1) \times 3$
$75 = (n - 1) \times 3$
$n - 1 = 25$
$n = 26.$
View full question & answer→Question 282 Marks
Find $a_{30} - a_{20}$ for the $A.P.$
$-9, -14, -19, -24, .....$
AnswerGiven,
$a_{30} - a_{20} = a + (30 - 1)d - (a + (20 - 1)d)$
($\therefore$$ a_n = a + (n - 1)d)$
$= a + 29d - a - 19d$
$= 10d$
$-9, -14, -19, -24, ....$
Common differecne (d) = Secomd term - First term
$= -14 - (-9)$
$= -14 + 9$
$d = -5$
Then $a_{30} - a_{20} = 10d$
$= 10 \times (-5)$
$= -50$
View full question & answer→Question 292 Marks
Find the sum of the following arithmetic progressions:
$50, 46, 42, .....$, to $10$ 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 progression is,
$50, 46, 42, ....$. to $10$ term.
First term $(a) = 50$
Common difference $(d) = 46 - 50 = -4$
$n^{th}$ term $= 10$
Then $\text{S}_{10}=\frac{10}{2}\big[2\times50+(10-1)(-4)\big]$
$= 5{100 - 9.4}$
$= 5{100 - 36}$
$= 5 \times 64$
$\therefore\ \text{S}_{10}=320$
View full question & answer→Question 302 Marks
Find the $12^{\text {th }}$ term from the end of the following arithmetic progressions: $3,8,13, \ldots . ., 253$.
AnswerIn the A.P. 3, 8, 13, ....., 253
First term (a) = 3
Common difference $( d )=8-3=5$
and Last term = 253
The $n ^{\text {th }}$ term from the last $=1-(n-1) d$
$12^{\text {th }}$ term from the last $=253-(12-1) \times 5=253-11 \times=253-55=198$
View full question & answer→Question 312 Marks
Find $a _{30}- a _{20}$ for the A.P.
$a, a+d, a+2 d, a+3 d, \ldots . .$
AnswerGiven,
$a_{30}-a_{20}=a+(30-1) d-(a+(20-1) d)\left(\therefore a_n=a+(n-1) d\right)$
$=a+29 d-a-19 d$
$=10 d$
In $A.P. a, a + d, a + 2d, a + 3d, .....$
$a$ is the first term and $d$ is the common difference
$\therefore$ $a_n = a + (n - 1)d$
$\therefore$ $a_{20} = a + (20 - 1)d = a + 19d$
and $a_{30} = a + (30 - 1)d = a + 29d$
$\therefore$ $a_{30} - a_{20} =a + 29d - a - 19d = 10d.$
View full question & answer→Question 322 Marks
The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles.
AnswerSuppose the angles of a triangle are (a - d), a, (a + d) such that, (a + d) > a > (a - d).
a - d + a + a + d = 180 [angle sum property]
⇒ 3a = 180
⇒ a = 60
Now, (a + d) = 2(a - d)
⇒ a + d = 2a - 2d
⇒ a = 3d
$\Rightarrow\ \text{d}=\frac{60}{3}=20$
Therefore, the three angles of a triangle are 40, 60, 80.
View full question & answer→Question 332 Marks
Show that $( a - b )^2,\left( a ^2+ b ^2\right)$ and $( a + b )^2$ are in $A.P$?
Answer$(a-b)^2,\left(a^2+b^2\right)$ and $(a+b)^2$ are in A.P.
If $2\left(a^2+b^2\right)=(a-b)^2+(a+b)^2$
If $2\left(a^2+b^2\right)=a^2+b^2-2 a b+a^2+b^2+2 a b$
If $2\left(a^2+b^2\right)=2 a^2+2 b^2=2\left(a^2+b^2\right)$
Which is true
Hence proved.
View full question & answer→Question 342 Marks
Find:
Which term in the A.P. $4,9,14, \ldots .$. is $254 ?$
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. 4, 9, 14, .....
First term (a) $=4$
Common difference $(d) = a_1 - a$
$= 9 - 4$
$= 5$
$n^{th}$ term $(a_n) = a + (n - 1)d$
Given $n ^{\text {th }}$ term is $254$
$4 + (n - 1)5 = 254$
$(n - 1).5 = 250$
$\text{n}-1=\frac{250}{5}=50$
$n = 50$
$\therefore$ $51^{st}$ term is $254.$
View full question & answer→Question 352 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$1.0, 1.7, 2.4, 3.1, .....$
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 $(d)$,
$1.0, 1.7, 2.4, 3.1, .....$
Here,
First term $(a) = 1.0$
$a_1 = 1.7$
$a_2 = 2.4$
Now, for the given to sequence to be an $A.P$.
Common difference $(d) = a_1- a = a_2 - a_1$
Here,
$a_1 - a = 1.7 - 1.0 = 0.7$
Also,
$a_2 - a_1 = 2.4 - 1.7 = 0.7$
Since $a_1 - a = a_2 - a_1$
Hence, the given sequence is an A.P. and its common difference is $d = 0.7.$
View full question & answer→Question 362 Marks
Find the indicated terms in the following sequences whose $n ^{\text {th }}$ terms are: $a_n=5 n-4 ; a_{12}$ and $a_{15}$.
Answer$a_n = 5n - 4$
$\therefore a_{12} = 5 \times 12 - 4 = 60 - 4 = 56$
$a_{15} = 5 \times 15 - 4 = 75 - 4 = 71.$
View full question & answer→Question 372 Marks
Find the common difference of the $A.P$. and write the next two terms:
$1.8, 2.0, 2.2, 2.4, .....$
Answer$1.8, 2.0, 2.2, 2.4, .....$
Let $a_1=1.8, a_2=2.0, a_3=2.2, a_4=2.4$
$\therefore$ $a_2 - a_1 = 2.0 - 1.8 = 0.2$
$a_3 - a_2 = 2.2 - 2.0 = 0.2$
$a_4 - a_3 = 2.4 - 2.2 = 0.2$
$\therefore$ Common difference $=0.2$
and text two terms will be,
$a_5 = 2.4 + 0.2 = 2.6$
$a_6 = 2.6 + 0.2 = 2.8$
$\therefore$ Next two terms are $2.6, 2.8.$
View full question & answer→Question 382 Marks
Write the $n^{th}$ ter, of the A.P. $\frac{1}{\text{m}},\frac{1+\text{m}}{\text{m}},\frac{1+2\text{m}}{\text{m}}, .....$
AnswerGiven: A.P. $\frac{1}{m}, \frac{1+ m }{ m }, \frac{1+2 m}{ m }, \ldots$.
We know that the $n^{\text {th }}$ term of an A.P. is given by
$a_n=a+(n-1) d$
In the given A.P..
$\text{a}=\frac{1}{\text{m}}$
$\text{d}=\frac{1+\text{m}}{\text{m}}-\frac{1}{\text{m}}=\frac{1+\text{m}-1}{\text{m}}=1$
Thus, the $n ^{\text {th }}$ term of the given A.P. is
$\text{a}_\text{n}=\frac{1}{\text{m}}+(\text{n}-1)1=\frac{1+(\text{n}-1)\text{m}}{\text{m}}$
View full question & answer→Question 392 Marks
Find the arithmetic progression whose third term is $16$ and seventh term exceeds its fifth term by $12$.
AnswerLet $a , a + d , a +2 d, a +3 d$,........ be the $A.P.$
$a_n = a + (n – 1)d$
But $a_3 = 16$
$a_7 - a_5 = 12$
Now $a_3 = a + (3 - 1)d = a + 2d$
$a_5 = a + (5 - 1)d = a + 4d$
and $a_7 = a + (7 - 1)d = a + 6d$
$\therefore$ $a + 2d = 16$
$\Rightarrow a = 16 - 2d$
and $a_7 - a_5 = a + 6d - a - 4d$
$\Rightarrow\ 12 = 2\text{d}\Rightarrow \text{d}=\frac{12}{2}=6$
$\therefore$ $a = 16 - 2d = 16 - 2 \times 6$
$\Rightarrow a = 16 - 12 = 4$
$\therefore$ Sequencing ($A.P$.) will be
$4, 10, 16, 22, .....$
View full question & answer→Question 402 Marks
If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
AnswerSince, (x + 1), 3x, (4x + 2) are in A.P.
So, we know difference of terms are common.
⇒ 3x - (x + 1) = (4x + 2) - 3x
⇒ 3x - x -1 = 4x + 2 - 3x
⇒ 2x - 1 = x + 2
⇒ x = 3
Hence, value of x is 3.
View full question & answer→Question 412 Marks
Find the $10^{\text {th }}$ term from the end of the A.P. $8,10,12, \ldots, 126$.
AnswerThe given A.P. is $8,10,12, \ldots . .126$
Here first term (a) = 8
Common difference $( d )=10-8=2$
and last term $(I)=126$
Now $n^{\text {th }}$ term from the last is $a_n=1-(n-1) d$
$a_{10} = 126 - (10 - 1) \times 2$
$= 126 - 9 \times 2$
$= 126 - 18$
$= 108.$
View full question & answer→Question 422 Marks
For the A.P. $-3,-7,-11, \ldots$, can we find $a_{30}-a_{20}$ without actually finding $a_{30}$ and $a_{20}$ ? Give reasons for your answer.
AnswerTrue.
$n^{\text {th }}$ term of an A.P., $a_n=a+(n-1) d$
$a_{30}=a+(30-1) d=a+29 d$.....(i)
and $a_{20}=a+(20-1) d=a+19 d$
Now, $a _{30}- a _{20}=( a +29 d)-( a +19 d)=10 d$
And from given A.P.
Common difference, $d =-7-(-3)=-7+3=-4$ $a_{30}-a_{20}=10(-4)=-40$ [from eq. (i)]
View full question & answer→Question 432 Marks
Find where 0 (zero) is a term of the A.P. $40,37,34,31, \ldots \ldots$.
Answer$A.P. 40, 37, 34, 31, .....$
Here,
$a = 40, d = -3$
Let $T_n = 0$
$T_n = a + (n - 1)d$
$\Rightarrow 0 = 40 + (n - 1)(-3)$
$\Rightarrow 0 = 40 - 3n + 3$
$\Rightarrow 3n = 43$
$\Rightarrow\ \text{n}=\frac{43}{3}$ which is in fraction
There is no term which is $0$.
View full question & answer→Question 442 Marks
Find the sum of,
The first 15 multiples of 8.
AnswerFirst 15 multiples of 8
are 8, 16, 24, 32, ...., 120
In which first term (a) = 8
Common difference (d) = 8
and Last term = 120
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_{15}=\frac{15}{2}[2\times8+(15-1)\times8]$
$=\frac{15}{2}[16+14\times8]$
$=\frac{15}{2}[16+112]$
$=\frac{15}{2}\times128=15\times64$
$=960$
View full question & answer→Question 452 Marks
Write the sequence with $n ^{\text {th }}$ term:
$a_n=9-5 n .$
Show the all of the above sequences form $A.P.$
AnswerThe $n ^{\text {th }}$ term an $=9-5 n$
Let $n=1,2,3,4,5, \ldots$, them
$a_1 = 9 - 5 \times 1 = 9 - 5 = 4$
$a_2 = 9 - 5 \times 2 = 9 - 10 = -1$
$a_3 = 9 - 5 \times 3 = 9 - 15 = -6$
$a_4 = 9 - 5 \times 4 = 9 - 20 = -11$
$a_5 = 9 - 5 \times 5 = 9 - 25 = -16$
We see that the sequence is
$4, -1, -6, -11, -16, ....$
Which is an $A.P$.
View full question & answer→Question 462 Marks
Find:Which term in the A.P. $21,42,63,84, \ldots .$. is 420 ?
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, $a_n=a+(n-1) d$.
A.P. $21,42,63,84, \ldots . .420$
Here first term $( a )=21$
Common difference $( d )=42-21=21$
Now $a_n=a+(n-1) d$
$\Rightarrow 420=21+(n-1) \times 21$
$\Rightarrow 420=21+21 n-21 \Rightarrow 420=21 n$
$\Rightarrow n =\frac{420}{21}=20$
$\therefore 420$ is the $20^{\text {th }}$ term.
View full question & answer→Question 472 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{2\text{n}-3}{6}$
AnswerGiven sequence is, $\text{a}_\text{n}=\frac{2\text{n}-3}{6}$
To write first five terms of given sequence we put $n = 1, 2, 3, 4, 5$. Then, we get,
$\text{a}_1=\frac{2.1-3}{6}=\frac{2-3}{6}=\frac{-1}{6}$
$\text{a}_2=\frac{2.2-3}{6}=\frac{4-3}{6}=\frac{1}{6}$
$\text{a}_3=\frac{2.3-3}{6}=\frac{6-3}{3}=\frac{1}{2}$
$\text{a}_4=\frac{2.4-3}{6}=\frac{8-3}{6}=\frac{5}{6}$
$\text{a}_5=\frac{2.5-3}{6}=\frac{10-3}{6}=\frac{7}{6}$
$\therefore$ The required first five terms of given sequence $\text{a}_\text{n}-\frac{2_\text{n}-3}{6}$ are $\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}.$
View full question & answer→Question 482 Marks
Find the sum:
$2 + 4 + 6 ..... + 200.$
Answer$2 + 4 + 6 + ..... + 200$
$a = 2, d = 4 - 2 = 2, l = 200 = a_n$
$\therefore\ \text{S}_{\text{n}}=\frac{\text{n}}{2}(\text{a}+\text{l})\text{ and a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$200=2+(\text{n}-1)2$
$198=(\text{n}-1)2$
$\text{n}-1=\frac{198}{2}=99$
$\text{n}=100$
$\text{S}_\text{n}=\frac{100}{2}(2+200)$
$=50\times202$
$=10100$
View full question & answer→Question 492 Marks
Find:$10^{th}$ term of the $A.P. -40, -15, 10, 35, .....$
AnswerGiven $A.P$. is
$-40, -15, 10, 35, .....$
First term (a) = -40
Common difference (d) = second term - first term
$= -15 - (-40)$
$= 40 - 15$
$= 25$
$n^{th}$ term of an $A.P. a_n = a + (n - 1)d$
$10^{th}$ term of $A.P. a_{10} = -40 + (10 - 1)25$
$= -40 + 9 \times 25$
$= -40 + 225$
$= 185$
View full question & answer→Question 502 Marks
Find the value of x for which (8x + 4), (6x - 2) and (2x + 7) are in A.P?
AnswerGiven,
8x + 4, 6x - 2, 2x + 7 are A.P.
If the numbers a, b, c are in A.P. then condition is 2b = a + c.
Then, 2(6x - 2) = 8x + 4 + 2x + 7
12x - 4 = 10x + 11
2x = 15
$\text{x}=\frac{15}{2}$
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