46 Arithmetic Progression Questions for Reference.

46 Arithmetic Progression Questions with Solutions

A complete question bank for Class 10 & 12 Board Prep provided by Omtex Classes.

Part 1: Basic Concepts and Nth Term

Question 1

Find the common difference of the AP: \( 5, 8, 11, 14, \dots \)

Solution: Common difference \( d = a_2 - a_1 \)
\( d = 8 - 5 = 3 \).
Answer: 3

Question 2

Find the 10th term of the AP: \( 2, 7, 12, \dots \)

Solution: Here, \( a = 2 \), \( d = 7 - 2 = 5 \), \( n = 10 \).
Using \( a_n = a + (n-1)d \):
\( a_{10} = 2 + (10-1)5 = 2 + 9(5) = 2 + 45 = 47 \).
Answer: 47

Question 3

Check if the list of numbers \( 2, 4, 8, 16, \dots \) forms an AP.

Solution: \( a_2 - a_1 = 4 - 2 = 2 \)
\( a_3 - a_2 = 8 - 4 = 4 \)
Since the difference is not constant, it is not an AP.

Question 4

Find the first term \( a \) if the common difference is -3 and the 10th term is -25.

Solution: Given \( d = -3, a_{10} = -25 \).
\( a_{10} = a + 9d \)
\( -25 = a + 9(-3) \)
\( -25 = a - 27 \Rightarrow a = 2 \).
Answer: 2

Question 5

Find the general term (\( n \)th term) of the AP: \( 13, 8, 3, -2, \dots \)

Solution: \( a = 13, d = 8 - 13 = -5 \).
\( a_n = 13 + (n-1)(-5) = 13 - 5n + 5 \).
Answer: \( a_n = 18 - 5n \)

Question 6

Which term of the AP \( 3, 8, 13, 18, \dots \) is 78?

Solution: \( a=3, d=5, a_n=78 \).
\( 78 = 3 + (n-1)5 \)
\( 75 = 5(n-1) \Rightarrow 15 = n - 1 \Rightarrow n = 16 \).
Answer: 16th term

Question 7

Find the 20th term from the last term of the AP: \( 3, 8, 13, \dots, 253 \).

Solution: Reverse the AP: \( 253, \dots, 13, 8, 3 \).
New \( a = 253 \), New \( d = -5 \).
\( a_{20} = 253 + (19)(-5) = 253 - 95 = 158 \).
Answer: 158

Question 8

Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution: \( a+2d = 5 \) (i)
\( a+6d = 9 \) (ii)
Subtract (i) from (ii): \( 4d = 4 \Rightarrow d = 1 \).
Substitute in (i): \( a + 2(1) = 5 \Rightarrow a = 3 \).
Answer: 3, 4, 5, 6, ...

Question 9

How many two-digit numbers are divisible by 3?

Solution: AP: \( 12, 15, \dots, 99 \).
\( a=12, d=3, a_n=99 \).
\( 99 = 12 + (n-1)3 \)
\( 87 = 3(n-1) \Rightarrow 29 = n - 1 \Rightarrow n = 30 \).
Answer: 30

Question 10

Find the middle term of the AP: \( 6, 13, 20, \dots, 216 \).

Solution: \( a=6, d=7, a_n=216 \).
\( 216 = 6 + (n-1)7 \Rightarrow 210 = 7(n-1) \Rightarrow n=31 \).
Middle term is \( \frac{31+1}{2} = 16 \)th term.
\( a_{16} = 6 + 15(7) = 6 + 105 = 111 \).
Answer: 111

Part 2: Finding Unknowns

Question 11

For what value of \( k \) are \( 2k, k+10, \) and \( 3k+2 \) in AP?

Solution: If \( a, b, c \) are in AP, \( 2b = a + c \).
\( 2(k+10) = 2k + (3k+2) \)
\( 2k + 20 = 5k + 2 \)
\( 18 = 3k \Rightarrow k = 6 \).
Answer: 6

Question 12

If \( x+2, 2x, 2x+3 \) are in AP, find \( x \).

Solution: \( 2(2x) = (x+2) + (2x+3) \)
\( 4x = 3x + 5 \Rightarrow x = 5 \).
Answer: 5

Question 13

Find \( a, b, c \) such that the numbers \( a, 7, b, 23, c \) are in AP.

Solution: \( a_2 = 7 \Rightarrow a+d=7 \). \( a_4 = 23 \Rightarrow a+3d=23 \).
Subtracting: \( 2d = 16 \Rightarrow d = 8 \).
\( a = -1, b = 15, c = 31 \).
Answer: a=-1, b=15, c=31

Question 14

Is 301 a term of the AP \( 5, 11, 17, 23, \dots \)?

Solution: \( a=5, d=6 \).
\( 301 = 5 + (n-1)6 \Rightarrow 296 = 6(n-1) \).
\( n-1 = 296/6 = 49.33 \).
Since \( n \) is not an integer, 301 is not a term.

Question 15

Find the value of \( p \) if the numbers \( 2p-1, 3p+1, 11 \) are in AP.

Solution: \( 2(3p+1) = (2p-1) + 11 \)
\( 6p + 2 = 2p + 10 \)
\( 4p = 8 \Rightarrow p = 2 \).
Answer: 2

Part 3: Sum of n Terms (Sn)

Question 16

Find the sum of the first 20 terms of the AP: \( 1, 4, 7, 10, \dots \)

Solution: \( a=1, d=3, n=20 \).
\( S_n = \frac{n}{2}[2a+(n-1)d] \)
\( S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[2 + 57] = 10(59) = 590 \).
Answer: 590

Question 17

Find the sum of the first 100 positive integers.

Solution: \( S_n = \frac{n(n+1)}{2} \)
\( S_{100} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \).
Answer: 5050

Question 18

Find the sum of: \( 34 + 32 + 30 + \dots + 10 \).

Solution: \( a=34, d=-2, l=10 \).
Find \( n \): \( 10 = 34 + (n-1)(-2) \Rightarrow -24 = -2(n-1) \Rightarrow n=13 \).
\( S_{13} = \frac{13}{2}(a+l) = \frac{13}{2}(34+10) = \frac{13}{2}(44) = 13 \times 22 = 286 \).
Answer: 286

Question 19

The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 + n \). Find the 2nd term.

Solution: \( a_1 = S_1 = 3(1)^2 + 1 = 4 \).
\( S_2 = 3(2)^2 + 2 = 14 \).
\( a_2 = S_2 - S_1 = 14 - 4 = 10 \).
Answer: 10

Question 20

How many terms of the AP \( 24, 21, 18, \dots \) must be taken so that their sum is 78?

Solution: \( a=24, d=-3, S_n=78 \).
\( 78 = \frac{n}{2}[48 + (n-1)(-3)] \Rightarrow 156 = n[48 - 3n + 3] \)
\( 156 = 51n - 3n^2 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0 \).
\( (n-4)(n-13) = 0 \). Both are valid.
Answer: 4 or 13

Question 21

Find the sum of the first 15 multiples of 8.

Solution: AP: \( 8, 16, \dots \). \( a=8, d=8, n=15 \).
\( S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}(16 + 112) = \frac{15}{2}(128) = 15 \times 64 = 960 \).
Answer: 960

Question 22

Find the sum of all odd numbers between 0 and 50.

Solution: AP: \( 1, 3, 5, \dots, 49 \). \( a=1, d=2, l=49 \).
\( 49 = 1 + (n-1)2 \Rightarrow 48 = 2(n-1) \Rightarrow n=25 \).
\( S_{25} = \frac{25}{2}(1+49) = \frac{25}{2}(50) = 625 \).
Answer: 625

Question 23

If the sum of first 7 terms is 49 and that of 17 terms is 289, find the sum of first \( n \) terms.

Solution: Given pattern: \( S_n = n^2 \).
(Proof: \( S_7=7^2, S_{17}=17^2 \)).
Answer: \( n^2 \)

Question 24

Find the sum of the first 40 positive integers divisible by 6.

Solution: \( a=6, d=6, n=40 \).
\( S_{40} = \frac{40}{2}[12 + 39(6)] = 20(12 + 234) = 20(246) = 4920 \).
Answer: 4920

Question 25

The first and the last terms of an AP are 17 and 350. If the common difference is 9, how many terms are there and what is their sum?

Solution: \( a=17, l=350, d=9 \).
\( 350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n=38 \).
\( S_{38} = \frac{38}{2}(17+350) = 19(367) = 6973 \).
Answer: n=38, Sum=6973

Part 4: Word Problems

Question 26

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Solution: AP: \( 5000, 5200, \dots, 7000 \).
\( a=5000, d=200, a_n=7000 \).
\( 7000 = 5000 + (n-1)200 \Rightarrow 2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n=11 \).
Year = \( 1995 + 10 = 2005 \).
Answer: 2005

Question 27

A sum of Rs 700 is to be used to give seven cash prizes. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

Solution: \( S_7 = 700, n=7, d=-20 \).
\( 700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 2a = 320 \Rightarrow a = 160 \).
Answer: 160, 140, 120, 100, 80, 60, 40

Question 28

In a school, students thought of planting 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 the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted?

Solution: Trees per class: \( 3 \times 1, 3 \times 2, \dots, 3 \times 12 \).
AP: \( 3, 6, 9, \dots, 36 \). \( n=12 \).
\( S_{12} = \frac{12}{2}(3+36) = 6(39) = 234 \).
Answer: 234 trees

Question 29

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed?

Solution: \( S_n = 200, a=20, d=-1 \).
\( 200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = n(41-n) = 41n - n^2 \).
\( n^2 - 41n + 400 = 0 \). Factors of 400 summing to 41 are 16 and 25.
If \( n=25, a_{25} = 20 - 24 = -4 \) (impossible). So \( n=16 \).
Answer: 16 rows

Question 30

Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the \( n \)th week, her weekly savings become Rs 20.75, find \( n \).

Solution: \( a=5, d=1.75, a_n=20.75 \).
\( 20.75 = 5 + (n-1)1.75 \)
\( 15.75 = (n-1)1.75 \Rightarrow n-1 = 9 \Rightarrow n=10 \).
Answer: 10

Part 5: Advanced & Properties

Question 31

Find three numbers in AP whose sum is 24 and whose product is 440.

Solution: Let terms be \( a-d, a, a+d \).
Sum: \( 3a = 24 \Rightarrow a = 8 \).
Product: \( (8-d)(8)(8+d) = 440 \)
\( 64 - d^2 = 55 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3 \).
Terms: \( 5, 8, 11 \).
Answer: 5, 8, 11

Question 32

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms.

Solution: \( (a+3d) + (a+7d) = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a+5d=12 \).
\( (a+5d) + (a+9d) = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a+7d=22 \).
Solving: \( 2d=10 \Rightarrow d=5 \). \( a = -13 \).
Answer: -13, -8, -3

Question 33

Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Solution: Let parts be \( a-d, a, a+d \). Sum \( 3a = 207 \Rightarrow a=69 \).
Smaller parts: \( (69-d)(69) = 4623 \).
\( 69-d = 67 \Rightarrow d=2 \).
Terms: \( 67, 69, 71 \).
Answer: 67, 69, 71

Question 34

Find the sum of all integers between 100 and 550 which are divisible by 9.

Solution: First term \( 108 \), Last term \( 549 \). \( d=9 \).
\( 549 = 108 + (n-1)9 \Rightarrow 441 = 9(n-1) \Rightarrow 49 = n-1 \Rightarrow n=50 \).
\( S_{50} = \frac{50}{2}(108+549) = 25(657) = 16425 \).
Answer: 16425

Question 35

If the \( m \)th term of an AP is \( 1/n \) and the \( n \)th term is \( 1/m \), show that the \( mn \)th term is 1.

Solution: \( a + (m-1)d = 1/n \) (1)
\( a + (n-1)d = 1/m \) (2)
Subtracting: \( (m-n)d = \frac{m-n}{mn} \Rightarrow d = \frac{1}{mn} \).
Substituting \( d \): \( a = \frac{1}{mn} \).
\( a_{mn} = \frac{1}{mn} + (mn-1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = 1 \).
Answer: 1 (Proved)

Question 36

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles.

Solution: Let angles be \( a-d, a, a+d \). Sum is 180.
\( 3a = 180 \Rightarrow a = 60^\circ \).
\( a+d = 2(a-d) \Rightarrow 60+d = 120 - 2d \Rightarrow 3d = 60 \Rightarrow d = 20 \).
Angles: \( 40^\circ, 60^\circ, 80^\circ \).
Answer: 40°, 60°, 80°

Question 37

Find the sum of the integers between 1 and 100 which are not divisible by 3.

Solution: Total sum (1 to 100) = 5050.
Sum of numbers divisible by 3 (\( 3, 6, \dots, 99 \)): \( n=33 \).
\( S_3 = \frac{33}{2}(3+99) = \frac{33}{2}(102) = 33 \times 51 = 1683 \).
Required Sum = \( 5050 - 1683 = 3367 \).
Answer: 3367

Question 38

If 9 times the 9th term of an AP is equal to 13 times the 13th term, find the 22nd term.

Solution: \( 9(a+8d) = 13(a+12d) \)
\( 9a + 72d = 13a + 156d \)
\( -4a = 84d \Rightarrow a = -21d \Rightarrow a + 21d = 0 \).
Since \( a_{22} = a+21d \), answer is 0.
Answer: 0

Question 39

If the sum of first \( m \) terms of an AP is same as the sum of its first \( n \) terms (\( m \neq n \)), find the sum of its first \( (m+n) \) terms.

Solution: \( S_m = S_n \Rightarrow \frac{m}{2}[2a+(m-1)d] = \frac{n}{2}[2a+(n-1)d] \).
Simplify: \( 2a(m-n) + d(m^2-m - n^2+n) = 0 \).
\( 2a(m-n) + d[(m-n)(m+n) - (m-n)] = 0 \).
Divide by \( m-n \): \( 2a + d(m+n-1) = 0 \).
\( S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d] = \frac{m+n}{2}(0) = 0 \).
Answer: 0

Question 40

The digits of a positive integer having three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

Solution: Digits \( a-d, a, a+d \). Sum \( 3a=15 \Rightarrow a=5 \).
Number: \( 100(5-d) + 10(5) + (5+d) \).
Reversed: \( 100(5+d) + 10(5) + (5-d) \).
Difference: \( -99(2d) = -594 \Rightarrow 198d = 594 \Rightarrow d = 3 \).
Digits: \( 8, 5, 2 \). Number: 852.
Answer: 852

Part 6: HOTS (Higher Order Thinking Skills)

Question 41

Determine \( k \) so that \( k^2 + 4k + 8, 2k^2 + 3k + 6, \) and \( 3k^2 + 4k + 4 \) are in AP.

Solution: \( 2b = a+c \).
\( 2(2k^2+3k+6) = (k^2+4k+8) + (3k^2+4k+4) \)
\( 4k^2+6k+12 = 4k^2+8k+12 \)
\( 6k = 8k \Rightarrow 2k = 0 \Rightarrow k = 0 \).
Answer: 0

Question 42

The sum of the first \( n \) terms of two APs are in the ratio \( (7n+1):(4n+27) \). Find the ratio of their \( m \)th terms.

Solution: Ratio of sums \( \frac{S_n}{S'_n} = \frac{2a+(n-1)d}{2a'+(n-1)d'} = \frac{7n+1}{4n+27} \).
To find ratio of \( m \)th term (\( \frac{a+(m-1)d}{a'+(m-1)d'} \)), replace \( n \) with \( 2m-1 \).
\( \frac{a_m}{a'_m} = \frac{7(2m-1)+1}{4(2m-1)+27} = \frac{14m-6}{8m+23} \).
Answer: \( (14m-6):(8m+23) \)

Question 43

Find the sum of all two-digit numbers which when divided by 4, yield 1 as remainder.

Solution: AP: \( 13, 17, \dots, 97 \).
\( 97 = 13 + (n-1)4 \Rightarrow 84 = 4(n-1) \Rightarrow 21 = n-1 \Rightarrow n=22 \).
\( S_{22} = \frac{22}{2}(13+97) = 11(110) = 1210 \).
Answer: 1210

Question 44

Solve the equation: \( 1 + 4 + 7 + 10 + \dots + x = 287 \).

Solution: \( a=1, d=3 \). Let total terms be \( n \).
\( 287 = \frac{n}{2}[2 + (n-1)3] \Rightarrow 574 = n(3n-1) \).
\( 3n^2 - n - 574 = 0 \).
Solving for \( n \): \( n = 14 \).
\( x = a_{14} = 1 + 13(3) = 1 + 39 = 40 \).
Answer: x = 40

Question 45

The 4th term of an AP is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the first term and common difference.

Solution: \( a+3d = 3a \Rightarrow 3d = 2a \) (i)
\( a+6d = 2(a+2d) + 1 \Rightarrow a+6d = 2a+4d+1 \Rightarrow 2d = a+1 \).
From (i), \( a = 1.5d \). Substitute: \( 2d = 1.5d + 1 \Rightarrow 0.5d = 1 \Rightarrow d = 2 \).
\( a = 1.5(2) = 3 \).
Answer: a=3, d=2

Question 46

If \( S_1, S_2, S_3 \) are the sums of \( n \) terms of three APs, the first term of each being 1 and the respective common differences being 1, 2, 3, prove that \( S_1 + S_3 = 2S_2 \).

Solution: \( S_1 = \frac{n}{2}[2 + (n-1)1] = \frac{n(n+1)}{2} \).
\( S_2 = \frac{n}{2}[2 + (n-1)2] = n^2 \).
\( S_3 = \frac{n}{2}[2 + (n-1)3] = \frac{n(3n-1)}{2} \).
\( S_1 + S_3 = \frac{n}{2}(n+1 + 3n-1) = \frac{n}{2}(4n) = 2n^2 = 2S_2 \).
Answer: Proved

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