Sum Of N Terms PYQs
Overview
This page provides comprehensive Class 10 Maths Sum Of N Terms PYQs | Arithmetic Progressions. Sum Of N Terms previous year questions for Class 10 Maths Arithmetic Progressions. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Q1
2025
00:00
Three numbers in A.P. have the sum 30. What is its middle term?
Let the three terms of the A.P. be $a-d$, $a$, $a+d$.
Sum of three terms = $(a-d)+a+(a+d)=3a$
Given sum = 30 ⇒ $3a=30$ ⇒ $a=10$
Middle term = 10 (Option b)
Q2
2023
00:00
If $a, b, c$ form an A.P. with common difference $d$, find the value of $a - 2b - c$.
Since $a, b, c$ are in A.P., we have $b=a+d$ and $c=a+2d$.
Substitute in the given expression:
$a - 2b - c = a - 2(a+d) - (a+2d)$
$= -2a - 4d$
$-2a - 4d$ (Option c)
Q3
NCERT, Delhi 2019
00:00
How many two-digit numbers are divisible by 3?
The smallest two-digit number divisible by 3 is 12.
The largest two-digit number divisible by 3 is 99.
The A.P. is: $12, 15, 18, \dots, 99$
Using $99 = 12+(n-1)3$, we get $n=30$
30 numbers
Q4
Delhi 2017
00:00
Find how many integers between 200 and 500 are divisible by 8.
The smallest number greater than 200 divisible by 8 is 208.
The largest number less than 500 divisible by 8 is 496.
The A.P. formed is $208, 216, 224, \dots, 496$.
Here $a=208$, $d=8$.
Using $496 = 208 + (n-1)8$, we get $n = 37$.
37 integers
Q5
2023
00:00
How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the last term is 62?
Given $a = -14$ and $a_5 = 2$.
Using $a_5 = a + 4d$, we get $2 = -14 + 4d$.
So $d = 4$.
Last term $l = 62$.
Using $62 = -14 + (n-1)4$, we get $n = 20$.
Number of terms = 20
Q6
2020
00:00
Find the sum of the first 100 natural numbers.
The natural numbers form the A.P. $1, 2, 3, \dots, 100$.
Here $a = 1$ and $n = 100$.
Using $S_n = \frac{n}{2}(a+l)$.
$S_{100} = \frac{100}{2}(1+100) = 50 \times 101$.
5050
Q7
Term II, 2021–22
00:00
Find the sum of first 20 terms of an A.P. in which $d = 5$ and $a_{20} = 135$.
Given $d = 5$ and $a_{20} = 135$.
Using $a_{20} = a + 19d$, we get $a = 40$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$S_{20} = 10(80 + 95)$.
1750
Q8
Term II, 2021–22
00:00
Find the sum of first 20 terms of an A.P. whose $n^{th}$ term is given by $a_n = 5 - 2n$.
Given $a_n = 5 - 2n$.
First term $a = a_1 = 3$.
Common difference $d = a_2 - a_1 = -2$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$S_{20} = 10(6 - 38)$.
$S_{20} = -320$
Q9
2018
00:00
Find the sum of first 8 multiples of 3.
The multiples of 3 form the A.P. $3, 6, 9, \dots$.
Here $a = 3$, $d = 3$ and $n = 8$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_8 = 4(27)$.
108
Q10
2020
00:00
Find the sum of first 16 terms of an A.P. whose 4th and 9th terms are −15 and −30 respectively.
Given $a_4 = a + 3d = -15$ and $a_9 = a + 8d = -30$.
Subtracting equations gives $5d = -15$ so $d = -3$.
Substituting back gives $a = -6$.
Using $S_{16} = \frac{16}{2}[2a + 15d]$.
$S_{16} = -456$
Q11
2020
00:00
Find the sum: $-5 + (-8) + (-11) + \dots + (-230)$.
The given series is an A.P. with first term $a = -5$ and common difference $d = -3$.
The last term is $l = -230$.
Using $l = a + (n-1)d$, we get $-230 = -5 + (n-1)(-3)$.
Solving, we get $n = 76$.
Using $S_n = \frac{n}{2}(a+l)$.
$S_{76} = \frac{76}{2}(-235)$.
$-8930$
Q12
Delhi 2017
00:00
Find the sum of $n$ terms of the series $(4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots$.
First term $a = 4 - \frac{1}{n}$.
Common difference $d = -\frac{1}{n}$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting values and simplifying.
$\dfrac{n(7n-1)}{2n}$
Q13
Delhi 2016
00:00
How many terms of the A.P. $18, 16, 14, \dots$ must be taken so that their sum is zero?
Here $a = 18$ and $d = -2$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Since sum is zero, we set $\frac{n}{2}[36 - 2(n-1)] = 0$.
Solving gives $n = 19$.
19 terms
Q14
Delhi 2016
00:00
How many terms of the A.P. $27, 24, 21, \dots$ should be taken so that their sum is zero?
Here $a = 27$ and $d = -3$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Setting the sum equal to zero and solving.
We get $n = 19$.
19 terms
Q15
Delhi 2017, 2024
00:00
If the sum of first $m$ terms of an A.P. is equal to the sum of its first $n$ terms $(m \neq n)$, show that the sum of its first $(m+n)$ terms is zero.
Let the first term be $a$ and common difference be $d$.
Given $S_m = S_n$.
Using sum formula and simplifying, we get $(m-n)[2a+(m+n-1)d] = 0$.
Since $m \neq n$, we get $2a+(m+n-1)d = 0$.
Using $S_{m+n} = \frac{m+n}{2}[2a+(m+n-1)d]$.
Hence proved that $S_{m+n} = 0$
Q16
2024
00:00
If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first 20 terms.
Let first term be $a$ and common difference be $d$.
From $S_7 = 49$, we get $a + 3d = 7$.
From $S_{17} = 289$, we get $a + 8d = 17$.
Solving gives $d = 2$ and $a = 1$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$S_{20} = 400$
Q17
2024
00:00
If the sum of first 14 terms of an A.P. is 1050 and the first term is 10, find the 20th term and the $n^{th}$ term.
Given $S_{14} = 1050$ and $a = 10$.
Using $S_{14} = \frac{14}{2}[2a + 13d]$.
Solving gives $d = 10$.
Then $a_{20} = a + 19d$.
General term $a_n = a + (n-1)d$.
$a_{20} = 200$, $a_n = 10n$
Q18
AI 2017, 2024
00:00
The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference.
Given $a = 5$, $l = 45$ and $S_n = 400$.
Using $S_n = \frac{n}{2}(a+l)$.
We get $n = 16$.
Using $l = a + (n-1)d$ to find $d$.
$n = 16$, $d = \frac{8}{3}$
Q19
2023
00:00
Find the sum of all integers between 50 and 500 which are divisible by 7.
The smallest number greater than 50 divisible by 7 is 56.
The largest number less than 500 divisible by 7 is 497.
The A.P. is $56, 63, 70, \dots, 497$.
Using $497 = 56 + (n-1)7$, we get $n = 64$.
Using $S_n = \frac{n}{2}(a+l)$.
17696
Q20
2020
00:00
In an A.P., $a = 54$, $d = -3$ and $a_n = 0$. Find $n$ and the sum of first $n$ terms.
Using $a_n = a + (n-1)d$.
Substituting values gives $0 = 54 - 3(n-1)$.
Solving gives $n = 19$.
Using $S_n = \frac{n}{2}(a+l)$.
$n = 19$, $S_n = 513$
Q21
2020
00:00
For an A.P., $a = 5$, $d = 3$ and $a_n = 50$. Find $n$ and the sum of first $n$ terms.
Using $a_n = a + (n-1)d$.
Substitute values: $50 = 5 + 3(n-1)$.
Solving gives $n = 16$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
$n = 16$, $S_n = 440$
Q22
Delhi 2017, 2019
00:00
If the $m^{th}$ term of an A.P. is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$, find the sum of its first $mn$ terms.
Let first term be $a$ and common difference be $d$.
Then $a + (m-1)d = \frac{1}{n}$ and $a + (n-1)d = \frac{1}{m}$.
Subtracting the equations gives $d = \frac{1}{mn}$.
Substituting back gives $a = \frac{1}{mn}$.
Using $S_{mn} = \frac{mn}{2}[2a + (mn-1)d]$.
$S_{mn} = \frac{mn+1}{2}$
Q23
NCERT, Delhi 2016
00:00
If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of first $n$ terms.
From earlier results, $a = 1$ and $d = 2$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Simplifying gives $S_n = n^2$.
$S_n = n^2$
Q24
Delhi 2016
00:00
How many terms of the A.P. $65, 60, 55, \dots$ should be taken so that their sum is zero?
Here $a = 65$ and $d = -5$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Setting the sum equal to zero and solving.
We get $n = 27$.
27 terms
Q25
AI 2016
00:00
If the ratio of the sums of first $n$ terms of two A.P.s is $(7n+1):(4n+27)$, find the ratio of their $n^{th}$ terms.
For large $n$, ratio of $n^{th}$ terms equals ratio of leading coefficients in sums.
Hence the ratio of $n^{th}$ terms is $(7n+1):(4n+27)$.
Ratio of $n^{th}$ terms = $(7n+1):(4n+27)$
Q26
AI 2016
00:00
The sums of first $n$ terms of three A.P.s are $S_1, S_2, S_3$. The first term of each is 1 and their common differences are 1, 2 and 3 respectively. Prove that $S_1 + S_3 = 2S_2$.
Using sum formula for each A.P.
$S_1 = \frac{n}{2}[2 + (n-1)]$.
$S_2 = \frac{n}{2}[2 + 2(n-1)]$.
$S_3 = \frac{n}{2}[2 + 3(n-1)]$.
Simplifying shows $S_1 + S_3 = 2S_2$.
Hence proved
Q27
Foreign 2016
00:00
The sums of first $n$ terms of three A.P.s are $S_1, S_2, S_3$. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that $S_1 + S_3 = 2S_2$.
Using sum formula for each A.P.
$S_1 = \frac{n}{2}[10 + 2(n-1)]$.
$S_2 = \frac{n}{2}[10 + 4(n-1)]$.
$S_3 = \frac{n}{2}[10 + 6(n-1)]$.
Simplifying shows $S_1 + S_3 = 2S_2$.
Hence proved
Q28
CBQ
00:00
In a charity run, the distances covered in successive rounds are 300 m, 350 m, 400 m, … for 10 rounds. Find the total distance covered.
The distances form an A.P. with $a = 300$, $d = 50$ and $n = 10$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_{10} = 5(600 + 450)$.
Total distance = 5250 m
Q29
2025
00:00
An A.P. consists of $n$ terms whose $n^{th}$ term is 4 and the common difference is 2. If the sum of $n$ terms is −14, find $n$. Also find the sum of the first 20 terms.
Given $a_n = 4$ and $d = 2$.
Using $a + (n-1)2 = 4$, we get $a = 6 - 2n$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$ and setting sum equal to −14.
Solving gives $n = 7$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$n = 7$, $S_{20} = 220$
Q30
2025
00:00
The sum of the first six terms of an A.P. is 42. The ratio of the 10th term to the 30th term is 1 : 3. Find the first term and the 13th term.
Let first term be $a$ and common difference be $d$.
Using $\frac{a+9d}{a+29d} = \frac{1}{3}$ gives $a = d$.
Given $S_6 = 42$, so $3(2a + 5d) = 42$.
Solving gives $a = 2$.
Using $a_{13} = a + 12d$.
First term = 2, 13th term = 26
Q31
2025
00:00
The sum of the 3rd and 7th terms of an A.P. is 6 and their product is 8. Find the sum of the first 16 terms of the A.P.
Let first term be $a$ and common difference be $d$.
Then $a_3 = a + 2d$ and $a_7 = a + 6d$.
Given $a_3 + a_7 = 6$ ⇒ $2a + 8d = 6$ ⇒ $a + 4d = 3$.
Also $(a+2d)(a+6d) = 8$.
Substituting $a = 3 - 4d$ gives $9 - 4d^2 = 8$.
So $d = \pm \frac{1}{2}$. Taking $d = \frac{1}{2}$, we get $a = 1$.
Using $S_{16} = \frac{16}{2}[2a + 15d]$.
$S_{16} = 76$
Q32
2020
00:00
Solve: $1 + 4 + 7 + 10 + \dots + x = 287$.
The series is an A.P. with $a = 1$ and $d = 3$.
Let number of terms be $n$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
We get $287 = \frac{n}{2}[2 + 3(n-1)]$.
Solving gives $n = 14$.
Thus $x = a_{14} = 1 + 13 \times 3$.
$x = 40$
Q33
2019
00:00
Find the sum of all odd numbers between 0 and 50.
Odd numbers between 0 and 50 are $1, 3, 5, \dots, 49$.
This is an A.P. with $a = 1$, $d = 2$ and $l = 49$.
Using $49 = 1 + (n-1)2$, we get $n = 25$.
Using $S_n = \frac{n}{2}(a+l)$.
625
Q34
Delhi 2019
00:00
The first term of an A.P. is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference.
Given $a = 3$, $l = 83$ and $S_n = 903$.
Using $S_n = \frac{n}{2}(a+l)$, we get $903 = 43n$.
So $n = 21$.
Using $l = a + (n-1)d$, we get $83 = 3 + 20d$.
$n = 21$, $d = 4$
Q35
AI 2019
00:00
Find the sum of all the two-digit numbers which leave remainder 2 when divided by 5.
Such numbers are $12, 17, 22, \dots, 97$.
This forms an A.P. with $a = 12$, $d = 5$ and $l = 97$.
Using $97 = 12 + (n-1)5$, we get $n = 18$.
Using $S_n = \frac{n}{2}(a+l)$.
981
Q36
Delhi 2017, Foreign 2016
00:00
The ratio of the sums of first $m$ and first $n$ terms of an A.P. is $m^2 : n^2$. Show that the ratio of its $m^{th}$ and $n^{th}$ terms is $(2m-1):(2n-1)$.
Let first term be $a$ and common difference be $d$.
Using sum formula, $S_m = \frac{m}{2}[2a + (m-1)d]$ and $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.
Simplifying gives $2a = d$.
Then $a_m = a + (m-1)d = \frac{(2m-1)d}{2}$ and similarly for $a_n$.
Hence $a_m : a_n = (2m-1):(2n-1)$
Q37
AI 2017
00:00
If the ratio of the sum of the first $n$ terms of two A.P.s is $(7n+1):(4n+27)$, find the ratio of their $n^{th}$ terms.
For large $n$, ratio of $n^{th}$ terms equals ratio of coefficients of $n^2$ in $S_n$.
Hence the ratio of $n^{th}$ terms is $(7n+1):(4n+27)$.
$(7n+1):(4n+27)$
Q38
2023
00:00
How many terms of the A.P. $45, 39, 33, \dots$ must be taken so that their sum is 180? Explain the double answer.
Here $a = 45$, $d = -6$ and $S_n = 180$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
We get $n(96 - 6n) = 360$.
Solving gives $n = 6$ or $n = 10$.
Double answer: $n = 6$ or $n = 10$
Q39
2020, 2018
00:00
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 the two middle terms is $7:15$. Find the numbers.
Let the numbers be $a-3d, a-d, a+d, a+3d$.
Given $4a = 32$ ⇒ $a = 8$.
Using given ratio and simplifying gives $d = 2$.
The numbers are $2, 6, 10, 14$
Q40
2019
00:00
Which term of the A.P. $-7, -12, -17, -22, \dots$ is $-82$? Is $-100$ a term of the A.P.? Give reason.
Here $a = -7$ and $d = -5$.
Using $a_n = a + (n-1)d$, we get $-82 = -7 - 5(n-1)$.
Solving gives $n = 16$.
For $-100$, $n$ is not an integer.
$-82$ is the 16th term; $-100$ is not a term
Q41
Delhi 2016
00:00
The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Let the numbers be $4-d, 4, 4+d$.
Using given condition on cubes gives $d^2 = 4$.
So $d = 2$.
The numbers are $2, 4, 6$
Q42
Term II, 2021–22
00:00
In an A.P., if $S_n = n(4n+1)$, find the A.P.
Using $a_n = S_n - S_{n-1}$.
We get $a_n = 8n - 3$.
So first term $a = 5$ and common difference $d = 8$.
The A.P. is $5, 13, 21, \dots$
Q43
Term II, 2021–22
00:00
Find the common difference of an A.P. whose first term is 10 and the sum of first 14 terms is 1505.
Using $S_{14} = \frac{14}{2}[2a + 13d]$.
Substituting $a = 10$ gives $7(20 + 13d) = 1505$.
Solving gives $d = 15$.
Common difference = 15
Three numbers in A.P. have the sum 30. What is its middle term?
Let the three terms of the A.P. be $a-d$, $a$, $a+d$.
Sum of three terms = $(a-d)+a+(a+d)=3a$
Given sum = 30 ⇒ $3a=30$ ⇒ $a=10$
Middle term = 10 (Option b)
Q2
2023
00:00
If $a, b, c$ form an A.P. with common difference $d$, find the value of $a - 2b - c$.
Since $a, b, c$ are in A.P., we have $b=a+d$ and $c=a+2d$.
Substitute in the given expression:
$a - 2b - c = a - 2(a+d) - (a+2d)$
$= -2a - 4d$
$-2a - 4d$ (Option c)
Q3
NCERT, Delhi 2019
00:00
How many two-digit numbers are divisible by 3?
The smallest two-digit number divisible by 3 is 12.
The largest two-digit number divisible by 3 is 99.
The A.P. is: $12, 15, 18, \dots, 99$
Using $99 = 12+(n-1)3$, we get $n=30$
30 numbers
Q4
Delhi 2017
00:00
Find how many integers between 200 and 500 are divisible by 8.
The smallest number greater than 200 divisible by 8 is 208.
The largest number less than 500 divisible by 8 is 496.
The A.P. formed is $208, 216, 224, \dots, 496$.
Here $a=208$, $d=8$.
Using $496 = 208 + (n-1)8$, we get $n = 37$.
37 integers
Q5
2023
00:00
How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the last term is 62?
Given $a = -14$ and $a_5 = 2$.
Using $a_5 = a + 4d$, we get $2 = -14 + 4d$.
So $d = 4$.
Last term $l = 62$.
Using $62 = -14 + (n-1)4$, we get $n = 20$.
Number of terms = 20
Q6
2020
00:00
Find the sum of the first 100 natural numbers.
The natural numbers form the A.P. $1, 2, 3, \dots, 100$.
Here $a = 1$ and $n = 100$.
Using $S_n = \frac{n}{2}(a+l)$.
$S_{100} = \frac{100}{2}(1+100) = 50 \times 101$.
5050
Q7
Term II, 2021–22
00:00
Find the sum of first 20 terms of an A.P. in which $d = 5$ and $a_{20} = 135$.
Given $d = 5$ and $a_{20} = 135$.
Using $a_{20} = a + 19d$, we get $a = 40$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$S_{20} = 10(80 + 95)$.
1750
Q8
Term II, 2021–22
00:00
Find the sum of first 20 terms of an A.P. whose $n^{th}$ term is given by $a_n = 5 - 2n$.
Given $a_n = 5 - 2n$.
First term $a = a_1 = 3$.
Common difference $d = a_2 - a_1 = -2$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$S_{20} = 10(6 - 38)$.
$S_{20} = -320$
Q9
2018
00:00
Find the sum of first 8 multiples of 3.
The multiples of 3 form the A.P. $3, 6, 9, \dots$.
Here $a = 3$, $d = 3$ and $n = 8$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_8 = 4(27)$.
108
Q10
2020
00:00
Find the sum of first 16 terms of an A.P. whose 4th and 9th terms are −15 and −30 respectively.
Given $a_4 = a + 3d = -15$ and $a_9 = a + 8d = -30$.
Subtracting equations gives $5d = -15$ so $d = -3$.
Substituting back gives $a = -6$.
Using $S_{16} = \frac{16}{2}[2a + 15d]$.
$S_{16} = -456$
Q11
2020
00:00
Find the sum: $-5 + (-8) + (-11) + \dots + (-230)$.
The given series is an A.P. with first term $a = -5$ and common difference $d = -3$.
The last term is $l = -230$.
Using $l = a + (n-1)d$, we get $-230 = -5 + (n-1)(-3)$.
Solving, we get $n = 76$.
Using $S_n = \frac{n}{2}(a+l)$.
$S_{76} = \frac{76}{2}(-235)$.
$-8930$
Q12
Delhi 2017
00:00
Find the sum of $n$ terms of the series $(4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots$.
First term $a = 4 - \frac{1}{n}$.
Common difference $d = -\frac{1}{n}$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting values and simplifying.
$\dfrac{n(7n-1)}{2n}$
Q13
Delhi 2016
00:00
How many terms of the A.P. $18, 16, 14, \dots$ must be taken so that their sum is zero?
Here $a = 18$ and $d = -2$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Since sum is zero, we set $\frac{n}{2}[36 - 2(n-1)] = 0$.
Solving gives $n = 19$.
19 terms
Q14
Delhi 2016
00:00
How many terms of the A.P. $27, 24, 21, \dots$ should be taken so that their sum is zero?
Here $a = 27$ and $d = -3$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Setting the sum equal to zero and solving.
We get $n = 19$.
19 terms
Q15
Delhi 2017, 2024
00:00
If the sum of first $m$ terms of an A.P. is equal to the sum of its first $n$ terms $(m \neq n)$, show that the sum of its first $(m+n)$ terms is zero.
Let the first term be $a$ and common difference be $d$.
Given $S_m = S_n$.
Using sum formula and simplifying, we get $(m-n)[2a+(m+n-1)d] = 0$.
Since $m \neq n$, we get $2a+(m+n-1)d = 0$.
Using $S_{m+n} = \frac{m+n}{2}[2a+(m+n-1)d]$.
Hence proved that $S_{m+n} = 0$
Q16
2024
00:00
If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first 20 terms.
Let first term be $a$ and common difference be $d$.
From $S_7 = 49$, we get $a + 3d = 7$.
From $S_{17} = 289$, we get $a + 8d = 17$.
Solving gives $d = 2$ and $a = 1$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$S_{20} = 400$
Q17
2024
00:00
If the sum of first 14 terms of an A.P. is 1050 and the first term is 10, find the 20th term and the $n^{th}$ term.
Given $S_{14} = 1050$ and $a = 10$.
Using $S_{14} = \frac{14}{2}[2a + 13d]$.
Solving gives $d = 10$.
Then $a_{20} = a + 19d$.
General term $a_n = a + (n-1)d$.
$a_{20} = 200$, $a_n = 10n$
Q18
AI 2017, 2024
00:00
The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference.
Given $a = 5$, $l = 45$ and $S_n = 400$.
Using $S_n = \frac{n}{2}(a+l)$.
We get $n = 16$.
Using $l = a + (n-1)d$ to find $d$.
$n = 16$, $d = \frac{8}{3}$
Q19
2023
00:00
Find the sum of all integers between 50 and 500 which are divisible by 7.
The smallest number greater than 50 divisible by 7 is 56.
The largest number less than 500 divisible by 7 is 497.
The A.P. is $56, 63, 70, \dots, 497$.
Using $497 = 56 + (n-1)7$, we get $n = 64$.
Using $S_n = \frac{n}{2}(a+l)$.
17696
Q20
2020
00:00
In an A.P., $a = 54$, $d = -3$ and $a_n = 0$. Find $n$ and the sum of first $n$ terms.
Using $a_n = a + (n-1)d$.
Substituting values gives $0 = 54 - 3(n-1)$.
Solving gives $n = 19$.
Using $S_n = \frac{n}{2}(a+l)$.
$n = 19$, $S_n = 513$
Q21
2020
00:00
For an A.P., $a = 5$, $d = 3$ and $a_n = 50$. Find $n$ and the sum of first $n$ terms.
Using $a_n = a + (n-1)d$.
Substitute values: $50 = 5 + 3(n-1)$.
Solving gives $n = 16$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
$n = 16$, $S_n = 440$
Q22
Delhi 2017, 2019
00:00
If the $m^{th}$ term of an A.P. is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$, find the sum of its first $mn$ terms.
Let first term be $a$ and common difference be $d$.
Then $a + (m-1)d = \frac{1}{n}$ and $a + (n-1)d = \frac{1}{m}$.
Subtracting the equations gives $d = \frac{1}{mn}$.
Substituting back gives $a = \frac{1}{mn}$.
Using $S_{mn} = \frac{mn}{2}[2a + (mn-1)d]$.
$S_{mn} = \frac{mn+1}{2}$
Q23
NCERT, Delhi 2016
00:00
If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of first $n$ terms.
From earlier results, $a = 1$ and $d = 2$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Simplifying gives $S_n = n^2$.
$S_n = n^2$
Q24
Delhi 2016
00:00
How many terms of the A.P. $65, 60, 55, \dots$ should be taken so that their sum is zero?
Here $a = 65$ and $d = -5$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
Setting the sum equal to zero and solving.
We get $n = 27$.
27 terms
Q25
AI 2016
00:00
If the ratio of the sums of first $n$ terms of two A.P.s is $(7n+1):(4n+27)$, find the ratio of their $n^{th}$ terms.
For large $n$, ratio of $n^{th}$ terms equals ratio of leading coefficients in sums.
Hence the ratio of $n^{th}$ terms is $(7n+1):(4n+27)$.
Ratio of $n^{th}$ terms = $(7n+1):(4n+27)$
Q26
AI 2016
00:00
The sums of first $n$ terms of three A.P.s are $S_1, S_2, S_3$. The first term of each is 1 and their common differences are 1, 2 and 3 respectively. Prove that $S_1 + S_3 = 2S_2$.
Using sum formula for each A.P.
$S_1 = \frac{n}{2}[2 + (n-1)]$.
$S_2 = \frac{n}{2}[2 + 2(n-1)]$.
$S_3 = \frac{n}{2}[2 + 3(n-1)]$.
Simplifying shows $S_1 + S_3 = 2S_2$.
Hence proved
Q27
Foreign 2016
00:00
The sums of first $n$ terms of three A.P.s are $S_1, S_2, S_3$. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that $S_1 + S_3 = 2S_2$.
Using sum formula for each A.P.
$S_1 = \frac{n}{2}[10 + 2(n-1)]$.
$S_2 = \frac{n}{2}[10 + 4(n-1)]$.
$S_3 = \frac{n}{2}[10 + 6(n-1)]$.
Simplifying shows $S_1 + S_3 = 2S_2$.
Hence proved
Q28
CBQ
00:00
In a charity run, the distances covered in successive rounds are 300 m, 350 m, 400 m, … for 10 rounds. Find the total distance covered.
The distances form an A.P. with $a = 300$, $d = 50$ and $n = 10$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_{10} = 5(600 + 450)$.
Total distance = 5250 m
Q29
2025
00:00
An A.P. consists of $n$ terms whose $n^{th}$ term is 4 and the common difference is 2. If the sum of $n$ terms is −14, find $n$. Also find the sum of the first 20 terms.
Given $a_n = 4$ and $d = 2$.
Using $a + (n-1)2 = 4$, we get $a = 6 - 2n$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$ and setting sum equal to −14.
Solving gives $n = 7$.
Using $S_{20} = \frac{20}{2}[2a + 19d]$.
$n = 7$, $S_{20} = 220$
Q30
2025
00:00
The sum of the first six terms of an A.P. is 42. The ratio of the 10th term to the 30th term is 1 : 3. Find the first term and the 13th term.
Let first term be $a$ and common difference be $d$.
Using $\frac{a+9d}{a+29d} = \frac{1}{3}$ gives $a = d$.
Given $S_6 = 42$, so $3(2a + 5d) = 42$.
Solving gives $a = 2$.
Using $a_{13} = a + 12d$.
First term = 2, 13th term = 26
Q31
2025
00:00
The sum of the 3rd and 7th terms of an A.P. is 6 and their product is 8. Find the sum of the first 16 terms of the A.P.
Let first term be $a$ and common difference be $d$.
Then $a_3 = a + 2d$ and $a_7 = a + 6d$.
Given $a_3 + a_7 = 6$ ⇒ $2a + 8d = 6$ ⇒ $a + 4d = 3$.
Also $(a+2d)(a+6d) = 8$.
Substituting $a = 3 - 4d$ gives $9 - 4d^2 = 8$.
So $d = \pm \frac{1}{2}$. Taking $d = \frac{1}{2}$, we get $a = 1$.
Using $S_{16} = \frac{16}{2}[2a + 15d]$.
$S_{16} = 76$
Q32
2020
00:00
Solve: $1 + 4 + 7 + 10 + \dots + x = 287$.
The series is an A.P. with $a = 1$ and $d = 3$.
Let number of terms be $n$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
We get $287 = \frac{n}{2}[2 + 3(n-1)]$.
Solving gives $n = 14$.
Thus $x = a_{14} = 1 + 13 \times 3$.
$x = 40$
Q33
2019
00:00
Find the sum of all odd numbers between 0 and 50.
Odd numbers between 0 and 50 are $1, 3, 5, \dots, 49$.
This is an A.P. with $a = 1$, $d = 2$ and $l = 49$.
Using $49 = 1 + (n-1)2$, we get $n = 25$.
Using $S_n = \frac{n}{2}(a+l)$.
625
Q34
Delhi 2019
00:00
The first term of an A.P. is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference.
Given $a = 3$, $l = 83$ and $S_n = 903$.
Using $S_n = \frac{n}{2}(a+l)$, we get $903 = 43n$.
So $n = 21$.
Using $l = a + (n-1)d$, we get $83 = 3 + 20d$.
$n = 21$, $d = 4$
Q35
AI 2019
00:00
Find the sum of all the two-digit numbers which leave remainder 2 when divided by 5.
Such numbers are $12, 17, 22, \dots, 97$.
This forms an A.P. with $a = 12$, $d = 5$ and $l = 97$.
Using $97 = 12 + (n-1)5$, we get $n = 18$.
Using $S_n = \frac{n}{2}(a+l)$.
981
Q36
Delhi 2017, Foreign 2016
00:00
The ratio of the sums of first $m$ and first $n$ terms of an A.P. is $m^2 : n^2$. Show that the ratio of its $m^{th}$ and $n^{th}$ terms is $(2m-1):(2n-1)$.
Let first term be $a$ and common difference be $d$.
Using sum formula, $S_m = \frac{m}{2}[2a + (m-1)d]$ and $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.
Simplifying gives $2a = d$.
Then $a_m = a + (m-1)d = \frac{(2m-1)d}{2}$ and similarly for $a_n$.
Hence $a_m : a_n = (2m-1):(2n-1)$
Q37
AI 2017
00:00
If the ratio of the sum of the first $n$ terms of two A.P.s is $(7n+1):(4n+27)$, find the ratio of their $n^{th}$ terms.
For large $n$, ratio of $n^{th}$ terms equals ratio of coefficients of $n^2$ in $S_n$.
Hence the ratio of $n^{th}$ terms is $(7n+1):(4n+27)$.
$(7n+1):(4n+27)$
Q38
2023
00:00
How many terms of the A.P. $45, 39, 33, \dots$ must be taken so that their sum is 180? Explain the double answer.
Here $a = 45$, $d = -6$ and $S_n = 180$.
Using $S_n = \frac{n}{2}[2a + (n-1)d]$.
We get $n(96 - 6n) = 360$.
Solving gives $n = 6$ or $n = 10$.
Double answer: $n = 6$ or $n = 10$
Q39
2020, 2018
00:00
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 the two middle terms is $7:15$. Find the numbers.
Let the numbers be $a-3d, a-d, a+d, a+3d$.
Given $4a = 32$ ⇒ $a = 8$.
Using given ratio and simplifying gives $d = 2$.
The numbers are $2, 6, 10, 14$
Q40
2019
00:00
Which term of the A.P. $-7, -12, -17, -22, \dots$ is $-82$? Is $-100$ a term of the A.P.? Give reason.
Here $a = -7$ and $d = -5$.
Using $a_n = a + (n-1)d$, we get $-82 = -7 - 5(n-1)$.
Solving gives $n = 16$.
For $-100$, $n$ is not an integer.
$-82$ is the 16th term; $-100$ is not a term
Q41
Delhi 2016
00:00
The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Let the numbers be $4-d, 4, 4+d$.
Using given condition on cubes gives $d^2 = 4$.
So $d = 2$.
The numbers are $2, 4, 6$
Q42
Term II, 2021–22
00:00
In an A.P., if $S_n = n(4n+1)$, find the A.P.
Using $a_n = S_n - S_{n-1}$.
We get $a_n = 8n - 3$.
So first term $a = 5$ and common difference $d = 8$.
The A.P. is $5, 13, 21, \dots$
Q43
Term II, 2021–22
00:00
Find the common difference of an A.P. whose first term is 10 and the sum of first 14 terms is 1505.
Using $S_{14} = \frac{14}{2}[2a + 13d]$.
Substituting $a = 10$ gives $7(20 + 13d) = 1505$.
Solving gives $d = 15$.
Common difference = 15