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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45 Advanced Arithmetic Progression Questions with Solutions

45 More Arithmetic Progression Questions with Solutions

Set 2: Advanced Practice for Class 10 & 12 Board Prep by Omtex Classes.

Part 1: Terms and General Concepts

Question 1

Find the next term of the AP: \( \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)

Solution: Simplifying the terms:
\(\sqrt{8} = 2\sqrt{2}\)
\(\sqrt{18} = 3\sqrt{2}\)
\(\sqrt{32} = 4\sqrt{2}\)
The sequence is \( 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots \)
Next term is \( 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50} \).
Answer: \(\sqrt{50}\)

Question 2

Find the common difference of the AP where \( a_{18} - a_{14} = 32 \).

Solution: \( a_{18} = a + 17d \) and \( a_{14} = a + 13d \).
\( (a + 17d) - (a + 13d) = 32 \)
\( 4d = 32 \Rightarrow d = 8 \).
Answer: 8

Question 3

Which term of the AP \( 21, 18, 15, \dots \) is -81?

Solution: \( a=21, d=-3, a_n=-81 \).
\( -81 = 21 + (n-1)(-3) \)
\( -102 = -3(n-1) \)
\( 34 = n - 1 \Rightarrow n = 35 \).
Answer: 35th term

Question 4

Is 0 a term of the AP \( 31, 28, 25, \dots \)?

Solution: \( a=31, d=-3 \).
Check if \( 0 = 31 + (n-1)(-3) \).
\( -31 = -3(n-1) \Rightarrow n-1 = 10.33 \).
Since \( n \) is not an integer, 0 is not a term.
Answer: No

Question 5

Find the 4th term from the end of the AP: \( -11, -8, -5, \dots, 49 \).

Solution: Reverse AP: \( 49, \dots, -8, -11 \).
New \( a = 49 \), New \( d = -3 \) (Original was +3).
\( a_4 = 49 + (3)(-3) = 49 - 9 = 40 \).
Answer: 40

Question 6

If the \( n \)th term of an AP is \( 5n - 3 \), find the difference between the 10th and 5th terms.

Solution: \( a_{10} = 5(10) - 3 = 47 \).
\( a_5 = 5(5) - 3 = 22 \).
Difference \( = 47 - 22 = 25 \).
Answer: 25

Question 7

Find the arithmetic mean between 13 and 19.

Solution: AM \( = \frac{a+b}{2} \)
\( = \frac{13+19}{2} = \frac{32}{2} = 16 \).
Answer: 16

Question 8

Find the number of terms in the AP: \( 18, 15\frac{1}{2}, 13, \dots, -47 \).

Solution: \( a=18, d = 15.5 - 18 = -2.5 \).
\( -47 = 18 + (n-1)(-2.5) \)
\( -65 = -2.5(n-1) \Rightarrow n-1 = 26 \Rightarrow n=27 \).
Answer: 27

Question 9

Determine the 10th term of the AP: \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, \dots \)

Solution: \( a = \frac{1}{m} \).
\( d = \frac{1+m}{m} - \frac{1}{m} = \frac{m}{m} = 1 \).
\( a_{10} = \frac{1}{m} + 9(1) = \frac{1}{m} + 9 = \frac{1+9m}{m} \).
Answer: \(\frac{1+9m}{m}\)

Question 10

Which term of the AP \( 5, 15, 25, \dots \) will be 130 more than its 31st term?

Solution: \( d=10 \).
\( a_n = a_{31} + 130 \)
\( a + (n-1)d = a + 30d + 130 \)
\( (n-1)10 = 30(10) + 130 \)
\( 10n - 10 = 300 + 130 = 430 \)
\( 10n = 440 \Rightarrow n = 44 \).
Answer: 44th term

Part 2: Sum of AP (Sn)

Question 11

Find the sum of the first 22 terms of the AP: \( 8, 3, -2, \dots \)

Solution: \( a=8, d=-5, n=22 \).
\( S_{22} = \frac{22}{2}[2(8) + 21(-5)] \)
\( = 11[16 - 105] = 11[-89] = -979 \).
Answer: -979

Question 12

If \( S_n = 5n^2 + 3n \), find the AP.

Solution: \( a_1 = S_1 = 5(1)^2 + 3(1) = 8 \).
\( S_2 = 5(2)^2 + 3(2) = 20 + 6 = 26 \).
\( a_2 = S_2 - S_1 = 26 - 8 = 18 \).
\( d = a_2 - a_1 = 18 - 8 = 10 \).
Answer: 8, 18, 28, ...

Question 13

How many terms of the AP \( 9, 17, 25, \dots \) must be taken to give a sum of 636?

Solution: \( a=9, d=8, S_n=636 \).
\( 636 = \frac{n}{2}[18 + (n-1)8] = \frac{n}{2}[8n + 10] = 4n^2 + 5n \).
\( 4n^2 + 5n - 636 = 0 \).
Using quadratic formula: \( n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} \)
\( n = \frac{-5 \pm 101}{8} \). Taking positive: \( n = 96/8 = 12 \).
Answer: 12

Question 14

Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

Solution: AP: \( 104, 108, \dots, 196 \).
\( 196 = 104 + (n-1)4 \Rightarrow 92 = 4(n-1) \Rightarrow n=24 \).
\( S_{24} = \frac{24}{2}(104 + 196) = 12(300) = 3600 \).
Answer: 3600

Question 15

Find the sum of the first 25 terms of an AP whose nth term is given by \( a_n = 7 - 3n \).

Solution: \( a_1 = 7 - 3(1) = 4 \).
\( a_{25} = 7 - 3(25) = 7 - 75 = -68 \).
\( S_{25} = \frac{25}{2}(4 - 68) = \frac{25}{2}(-64) = 25(-32) = -800 \).
Answer: -800

Question 16

The sum of the first 6 terms of an AP is 36 and the sum of the first 16 terms is 256. Find the sum of the first 10 terms.

Solution: Note the pattern: \( S_n = n^2 \).
\( S_6 = 6^2 = 36 \), \( S_{16} = 16^2 = 256 \).
Therefore, \( S_{10} = 10^2 = 100 \).
Answer: 100

Question 17

Find the sum: \( (-5) + (-8) + (-11) + \dots + (-230) \).

Solution: \( a=-5, d=-3, l=-230 \).
\( -230 = -5 + (n-1)(-3) \Rightarrow -225 = -3(n-1) \Rightarrow n=76 \).
\( S_{76} = \frac{76}{2}(-5 - 230) = 38(-235) = -8930 \).
Answer: -8930

Question 18

If the sum of \( n \) terms of an AP is \( 2n^2 + 5n \), find the common difference.

Solution: Comparing with \( S_n = \frac{d}{2}n^2 + (a - \frac{d}{2})n \), coefficient of \( n^2 \) is \( d/2 \).
\( d/2 = 2 \Rightarrow d = 4 \).
Alternatively: \( S_1 = 7 = a \), \( S_2 = 18 \). \( a_2 = 11 \). \( d = 4 \).
Answer: 4

Part 3: Finding Unknown Variables

Question 19

Find \( x \) if \( 2x, x+10, 3x+2 \) are in AP.

Solution: \( 2(x+10) = 2x + (3x+2) \)
\( 2x + 20 = 5x + 2 \)
\( 18 = 3x \Rightarrow x = 6 \).
Answer: 6

Question 20

If \( k+9, 2k-1, \) and \( 2k+7 \) are in AP, find \( k \).

Solution: \( 2(2k-1) = (k+9) + (2k+7) \)
\( 4k - 2 = 3k + 16 \)
\( k = 18 \).
Answer: 18

Question 21

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

Solution: \( 2(2k^2+3k+6) = (4k+8) + (3k^2+4k+4) \)
\( 4k^2+6k+12 = 3k^2+8k+12 \)
\( k^2 - 2k = 0 \Rightarrow k(k-2) = 0 \).
Answer: 0 or 2

Question 22

Find the value of \( a \) and \( b \) given that the numbers \( 2, a, 10, b \) are in AP.

Solution: Common difference must be constant.
\( a-2 = 10-a \Rightarrow 2a=12 \Rightarrow a=6 \).
\( d = 6-2=4 \).
\( b = 10+4 = 14 \).
Answer: a=6, b=14

Question 23

If \( \frac{4}{5}, k, 2 \) are in AP, find \( k \).

Solution: \( 2k = \frac{4}{5} + 2 = \frac{14}{5} \).
\( k = \frac{7}{5} \).
Answer: 7/5

Part 4: Real Life Word Problems

Question 24

A man repays a loan of Rs 3250 by paying Rs 20 in the first month and then increasing the payment by Rs 15 every month. How long will it take him to clear the loan?

Solution: \( S_n = 3250, a=20, d=15 \).
\( 3250 = \frac{n}{2}[40 + (n-1)15] \)
\( 6500 = n(40 + 15n - 15) = n(25 + 15n) = 5n(5 + 3n) \).
\( 1300 = 5n + 3n^2 \Rightarrow 3n^2 + 5n - 1300 = 0 \).
Factors of \( 3 \times -1300 = -3900 \) summing to 5 are 65 and -60.
\( n = 20 \) or \( n = -65/3 \).
Answer: 20 months

Question 25

The taxi fare after each km when the fare is Rs 15 for the first km and rises by Rs 8 for each additional km. Write the AP and find the fare for 15 km.

Solution: AP: \( 15, 23, 31, \dots \).
\( a=15, d=8, n=15 \).
\( a_{15} = 15 + 14(8) = 15 + 112 = 127 \).
Answer: Rs 127

Question 26

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming uniform increase, find production in the 1st year.

Solution: \( a_3 = 600 \Rightarrow a+2d=600 \).
\( a_7 = 700 \Rightarrow a+6d=700 \).
Subtracting: \( 4d = 100 \Rightarrow d = 25 \).
\( a + 50 = 600 \Rightarrow a = 550 \).
Answer: 550 sets

Question 27

In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m apart in a straight line. There are 10 potatoes. Find the total distance run by a competitor.

Solution: Distances run: \( 2(5), 2(5+3), 2(5+6), \dots \)
AP: \( 10, 16, 22, \dots \). \( n=10 \).
\( S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = 370 \).
Answer: 370 m

Question 28

A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, etc. What is the total length of such a spiral made up of 13 consecutive semicircles?

Solution: Perimeter \( \pi r \). \( r_1=0.5, r_2=1.0 \).
AP of lengths: \( 0.5\pi, 1.0\pi, 1.5\pi, \dots \).
\( a=0.5\pi, d=0.5\pi, n=13 \).
\( S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi) \).
Using \( \pi = 22/7 \): \( \frac{91}{2} \times \frac{22}{7} = 13 \times 11 = 143 \).
Answer: 143 cm

Part 5: Advanced & Properties

Question 29

Divide 32 into four parts which are in AP such that the product of extremes is to the product of means is 7:15.

Solution: Let parts be \( a-3d, a-d, a+d, a+3d \). Sum = 32 \(\Rightarrow 4a=32 \Rightarrow a=8 \).
\( \frac{(8-3d)(8+3d)}{(8-d)(8+d)} = \frac{7}{15} \).
\( \frac{64-9d^2}{64-d^2} = \frac{7}{15} \).
\( 15(64-9d^2) = 7(64-d^2) \Rightarrow 960 - 135d^2 = 448 - 7d^2 \).
\( 512 = 128d^2 \Rightarrow d^2 = 4 \Rightarrow d = 2 \).
Parts: \( 2, 6, 10, 14 \).
Answer: 2, 6, 10, 14

Question 30

If \( p, q, r \) are in AP, prove that \( p^3 + r^3 + 6pqr = 8q^3 \).

Solution: Since in AP, \( p+r = 2q \).
Cube both sides: \( (p+r)^3 = (2q)^3 \).
\( p^3 + r^3 + 3pr(p+r) = 8q^3 \).
Substitute \( p+r=2q \): \( p^3 + r^3 + 3pr(2q) = 8q^3 \).
\( p^3 + r^3 + 6pqr = 8q^3 \).
Answer: Proved

Question 31

Find the sum of all two digit numbers which leave remainder 1 when divided by 3.

Solution: AP: \( 10, 13, 16, \dots, 97 \).
\( 97 = 10 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow n=30 \).
\( S_{30} = \frac{30}{2}(10+97) = 15(107) = 1605 \).
Answer: 1605

Question 32

If \( a^2, b^2, c^2 \) are in AP, prove that \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are in AP.

Solution: Given \( 2b^2 = a^2 + c^2 \).
Check AP condition: \( \frac{1}{c+a} - \frac{1}{b+c} = \frac{1}{a+b} - \frac{1}{c+a} \).
LHS = \( \frac{b+c-c-a}{(c+a)(b+c)} = \frac{b-a}{(c+a)(b+c)} \).
RHS = \( \frac{c+a-a-b}{(a+b)(c+a)} = \frac{c-b}{(a+b)(c+a)} \).
Cross multiply \( \frac{b-a}{b+c} = \frac{c-b}{a+b} \Rightarrow b^2-a^2 = c^2-b^2 \Rightarrow 2b^2 = a^2+c^2 \).
Since this is true, the terms are in AP.
Answer: Proved

Question 33

The sum of \( n, 2n, 3n \) terms of an AP are \( S_1, S_2, S_3 \) respectively. Prove that \( S_3 = 3(S_2 - S_1) \).

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

Question 34

Find the common difference of an AP whose first term is 5 and the sum of the first four terms is half the sum of the next four terms.

Solution: \( S_4 = \frac{1}{2}(S_8 - S_4) \Rightarrow 2S_4 = S_8 - S_4 \Rightarrow 3S_4 = S_8 \).
\( 3 \times \frac{4}{2}[2(5)+3d] = \frac{8}{2}[2(5)+7d] \).
\( 6[10+3d] = 4[10+7d] \)
\( 60 + 18d = 40 + 28d \)
\( 20 = 10d \Rightarrow d = 2 \).
Answer: 2

Part 6: HOTS (Higher Order Thinking Skills)

Question 35

Which term of the AP \( 121, 117, 113, \dots \) is its first negative term?

Solution: \( a=121, d=-4 \).
We need \( a_n < 0 \).
\( 121 + (n-1)(-4) < 0 \)
\( 121 - 4n + 4 < 0 \Rightarrow 125 < 4n \Rightarrow n > 31.25 \).
First integer is 32.
Answer: 32nd term

Question 36

If the roots of the cubic equation \( x^3 - 12x^2 + 39x - 28 = 0 \) are in AP, find them.

Solution: Let roots be \( a-d, a, a+d \).
Sum of roots = \( -(-12)/1 = 12 \).
\( (a-d)+a+(a+d) = 12 \Rightarrow 3a = 12 \Rightarrow a = 4 \).
Product of roots = \( -(-28)/1 = 28 \).
\( (4-d)(4)(4+d) = 28 \Rightarrow 16-d^2 = 7 \Rightarrow d^2 = 9 \Rightarrow d=3 \).
Roots: \( 1, 4, 7 \).
Answer: 1, 4, 7

Question 37

The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

Solution: \( a_3 + a_7 = 6 \Rightarrow 2a + 8d = 6 \Rightarrow a + 4d = 3 \Rightarrow a = 3-4d \).
\( a_3 \times a_7 = 8 \Rightarrow (a+2d)(a+6d) = 8 \).
Substitute \( a \): \( (3-2d)(3+2d) = 8 \).
\( 9 - 4d^2 = 8 \Rightarrow 4d^2 = 1 \Rightarrow d = \pm 1/2 \).
Case 1: \( d=1/2, a=1 \). \( S_{16} = 8[2 + 15(0.5)] = 76 \).
Case 2: \( d=-1/2, a=5 \). \( S_{16} = 8[10 - 7.5] = 20 \).
Answer: 76 or 20

Question 38

If \( S_n \) denotes the sum of first \( n \) terms of an AP, prove that \( S_{12} = 3(S_8 - S_4) \).

Solution: \( S_8 - S_4 = \frac{8}{2}(2a+7d) - \frac{4}{2}(2a+3d) \)
\( = 4(2a+7d) - 2(2a+3d) = 8a+28d-4a-6d = 4a+22d \).
Multiply by 3: \( 12a + 66d \).
\( S_{12} = \frac{12}{2}(2a+11d) = 6(2a+11d) = 12a + 66d \).
Answer: Proved

Question 39

Solve for \( x \): \( -4 + (-1) + 2 + \dots + x = 437 \).

Solution: \( a=-4, d=3 \).
\( 437 = \frac{n}{2}[2(-4) + (n-1)3] \).
\( 874 = n(-8 + 3n - 3) = 3n^2 - 11n \).
\( 3n^2 - 11n - 874 = 0 \). Using quadratic formula, \( n=19 \).
\( x = a_{19} = -4 + 18(3) = 50 \).
Answer: 50

Question 40

Find the AP if the 4th term is 18 and the difference of the 9th and 15th term is 30.

Solution: \( a_{15} - a_9 = 30 \Rightarrow 6d = 30 \Rightarrow d = 5 \).
\( a_4 = 18 \Rightarrow a + 3(5) = 18 \Rightarrow a = 3 \).
Answer: 3, 8, 13, ...

Question 41

Find the sum of all multiples of 7 lying between 500 and 900.

Solution: First multiple: 504. Last multiple: 896.
\( 896 = 504 + (n-1)7 \Rightarrow 392 = 7(n-1) \Rightarrow 56 = n-1 \Rightarrow n=57 \).
\( S_{57} = \frac{57}{2}(504+896) = \frac{57}{2}(1400) = 39900 \).
Answer: 39900

Question 42

If the sum of \( m \) terms of an AP is \( n \) and the sum of \( n \) terms is \( m \), then find the sum of \( (m+n) \) terms.

Solution: \( 2a + (m-1)d = \frac{2n}{m} \) (1)
\( 2a + (n-1)d = \frac{2m}{n} \) (2)
Subtracting: \( d(m-n) = \frac{2n}{m} - \frac{2m}{n} = \frac{2(n^2-m^2)}{mn} \).
\( d = -\frac{2(m+n)}{mn} \).
Substitute to find \( S_{m+n} = -(m+n) \).
Answer: -(m+n)

Question 43

A club consists of members whose ages are in AP, the common difference being 3 months. If the youngest member is 7 years old and the sum of the ages of all members is 250 years, find the number of members.

Solution: Convert all to years. \( d = 3/12 = 0.25 \). \( a=7 \). \( S_n = 250 \).
\( 250 = \frac{n}{2}[14 + (n-1)0.25] \).
\( 500 = n(14 + 0.25n - 0.25) = 13.75n + 0.25n^2 \).
\( 0.25n^2 + 13.75n - 500 = 0 \). Multiply by 4:
\( n^2 + 55n - 2000 = 0 \).
Factors: \( (n+80)(n-25)=0 \).
Answer: 25 members

Question 44

Find the sum of odd integers from 1 to 2001.

Solution: \( a=1, l=2001, d=2 \).
\( 2001 = 1 + (n-1)2 \Rightarrow 2000 = 2(n-1) \Rightarrow n=1001 \).
\( S_n = \frac{1001}{2}(1+2001) = 1001 \times 1001 = 1002001 \).
Answer: 1002001

Question 45

The ratio of the sum of \( m \) and \( n \) terms of an AP is \( m^2 : n^2 \). Show that the ratio of \( m \)th and \( n \)th term is \( 2m-1 : 2n-1 \).

Solution: \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \Rightarrow \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]} = \frac{m^2}{n^2} \).
\( \frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n} \).
This holds if \( d=2a \).
Ratio of terms: \( \frac{a+(m-1)2a}{a+(n-1)2a} = \frac{a(1+2m-2)}{a(1+2n-2)} = \frac{2m-1}{2n-1} \).
Answer: Proved

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20 Additional Arithmetic Progression Questions with Solutions

20 Arithmetic Progression Questions (Set 5)

Final Set: Complete your mastery with Omtex Classes.

Part 1: Concepts and Calculations

Question 1

Find the 30th term of the AP: \( 10, 7, 4, \dots \)

Solution: \( a = 10 \), \( d = 7 - 10 = -3 \).
\( a_{30} = 10 + (30-1)(-3) \)
\( a_{30} = 10 + 29(-3) = 10 - 87 = -77 \).
Answer: -77

Question 2

Find the number of terms in the AP: \( -1, -5/6, -2/3, \dots, 10/3 \).

Solution: \( a = -1 \).
\( d = -5/6 - (-1) = -5/6 + 6/6 = 1/6 \).
\( a_n = 10/3 \Rightarrow \frac{10}{3} = -1 + (n-1)\frac{1}{6} \).
\( \frac{13}{3} = \frac{n-1}{6} \Rightarrow 26 = n-1 \Rightarrow n = 27 \).
Answer: 27

Question 3

Which term of the AP \( 3, 15, 27, 39, \dots \) will be 132 more than its 54th term?

Solution: \( d = 12 \).
\( a_n = a_{54} + 132 \)
\( a + (n-1)d = a + 53d + 132 \)
\( (n-1)12 = 53(12) + 132 \). (Divide by 12)
\( n-1 = 53 + 11 = 64 \Rightarrow n = 65 \).
Answer: 65th term

Question 4

Find the value of \( x \) for which \( 2x, x+10, 3x+2 \) are three consecutive terms of an AP.

Solution: \( 2(x+10) = 2x + (3x+2) \)
\( 2x + 20 = 5x + 2 \)
\( 18 = 3x \Rightarrow x = 6 \).
Answer: 6

Question 5

If \( 1/x, 1/y, 1/z \) are in AP, prove that \( y = \frac{2xz}{x+z} \).

Solution: Since in AP, \( 2(\frac{1}{y}) = \frac{1}{x} + \frac{1}{z} \).
\( \frac{2}{y} = \frac{z+x}{xz} \).
Inverting both sides: \( \frac{y}{2} = \frac{xz}{x+z} \Rightarrow y = \frac{2xz}{x+z} \).
Answer: Proved

Question 6

Find the sum of all natural numbers between 100 and 500 which are divisible by 8.

Solution: First multiple 104, Last multiple 496.
\( 496 = 104 + (n-1)8 \Rightarrow 392 = 8(n-1) \Rightarrow 49 = n-1 \Rightarrow n = 50 \).
\( S_{50} = \frac{50}{2}(104 + 496) = 25(600) = 15000 \).
Answer: 15000

Question 7

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution: \( S_{14} = 1050, a=10 \).
\( 1050 = \frac{14}{2}[2(10) + 13d] \Rightarrow 1050 = 7[20 + 13d] \).
\( 150 = 20 + 13d \Rightarrow 130 = 13d \Rightarrow d = 10 \).
\( a_{20} = 10 + 19(10) = 200 \).
Answer: 200

Question 8

Find the sum of the series: \( 72 + 70 + 68 + \dots + 40 \).

Solution: \( a=72, d=-2, l=40 \).
\( 40 = 72 + (n-1)(-2) \Rightarrow -32 = -2(n-1) \Rightarrow 16 = n-1 \Rightarrow n=17 \).
\( S_{17} = \frac{17}{2}(72+40) = \frac{17}{2}(112) = 17 \times 56 = 952 \).
Answer: 952

Question 9

If the sum of \( n \) terms of an AP is \( S_n = 3n^2 + 5n \), find the AP.

Solution: \( S_1 = 3(1) + 5 = 8 \) (First term \( a=8 \)).
\( S_2 = 3(4) + 10 = 22 \).
\( a_2 = S_2 - S_1 = 22 - 8 = 14 \).
\( d = 14 - 8 = 6 \).
Answer: 8, 14, 20, ...

Question 10

Divide 15 into three parts which are in AP and such that the sum of their squares is 83.

Solution: Parts: \( a-d, a, a+d \). Sum \( 3a=15 \Rightarrow a=5 \).
Squares: \( (5-d)^2 + 25 + (5+d)^2 = 83 \).
\( 25 - 10d + d^2 + 25 + 25 + 10d + d^2 = 83 \).
\( 75 + 2d^2 = 83 \Rightarrow 2d^2 = 8 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Parts: \( 3, 5, 7 \).
Answer: 3, 5, 7

Part 2: Advanced Problems

Question 11

The angles of a quadrilateral are in AP whose common difference is 10°. Find the angles.

Solution: Let angles be \( a, a+10, a+20, a+30 \).
Sum of angles of quadrilateral = 360°.
\( 4a + 60 = 360 \Rightarrow 4a = 300 \Rightarrow a = 75 \).
Angles: \( 75^\circ, 85^\circ, 95^\circ, 105^\circ \).
Answer: 75°, 85°, 95°, 105°

Question 12

If the sum of \( p \) terms of an AP is \( q \) and the sum of \( q \) terms is \( p \), find the sum of \( (p+q) \) terms.

Solution: \( S_p = q \) and \( S_q = p \).
Using the standard result for this specific pattern: \( S_{p+q} = -(p+q) \).
Answer: -(p+q)

Question 13

Find the middle term(s) of the AP: \( 7, 13, 19, \dots, 241 \).

Solution: \( a=7, d=6, a_n=241 \).
\( 241 = 7 + (n-1)6 \Rightarrow 234 = 6(n-1) \Rightarrow 39 = n-1 \Rightarrow n=40 \).
Since \( n \) is even, there are two middle terms: \( n/2 = 20 \) and \( (n/2)+1 = 21 \).
\( a_{20} = 7 + 19(6) = 121 \).
\( a_{21} = 7 + 20(6) = 127 \).
Answer: 121 and 127

Question 14

Solve the equation: \( -4 + (-1) + 2 + \dots + x = 437 \).

Solution: \( a=-4, d=3 \). \( S_n = 437 \).
\( 437 = \frac{n}{2}[2(-4) + (n-1)3] \)
\( 874 = n(-8 + 3n - 3) = n(3n-11) = 3n^2 - 11n \).
\( 3n^2 - 11n - 874 = 0 \). Using quadratic formula, \( n=19 \).
\( x = a_{19} = -4 + 18(3) = 50 \).
Answer: 50

Question 15

Find the sum of all odd numbers between 10 and 200.

Solution: AP: \( 11, 13, \dots, 199 \).
\( 199 = 11 + (n-1)2 \Rightarrow 188 = 2(n-1) \Rightarrow 94 = n-1 \Rightarrow n=95 \).
\( S_{95} = \frac{95}{2}(11+199) = \frac{95}{2}(210) = 95 \times 105 = 9975 \).
Answer: 9975

Question 16

The sums of \( n \) terms of two APs are in the ratio \( (3n+8):(7n+15) \). Find the ratio of their 12th terms.

Solution: To find ratio of \( m \)th term, replace \( n \) with \( 2m-1 \).
Here \( m=12 \), so \( n = 2(12)-1 = 23 \).
Ratio = \( \frac{3(23)+8}{7(23)+15} = \frac{69+8}{161+15} = \frac{77}{176} \).
Dividing by 11: \( 7/16 \).
Answer: 7:16

Question 17

Determine the 2nd term of an AP whose 6th term is 12 and 8th term is 22.

Solution: \( a+5d=12 \) and \( a+7d=22 \).
Subtracting: \( 2d = 10 \Rightarrow d=5 \).
\( a + 25 = 12 \Rightarrow a = -13 \).
\( a_2 = a + d = -13 + 5 = -8 \).
Answer: -8

Question 18

Jaspal saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If he continues to save in this manner, in how many months will he save Rs 2000?

Solution: \( a=32, d=4, S_n=2000 \).
\( 2000 = \frac{n}{2}[64 + (n-1)4] \).
\( 4000 = n(60+4n) = 4n^2 + 60n \).
\( 4n^2 + 60n - 4000 = 0 \). Divide by 4: \( n^2 + 15n - 1000 = 0 \).
Factors of -1000 summing to 15: 40 and -25.
\( (n+40)(n-25) = 0 \). Time cannot be negative.
Answer: 25 months

Question 19

If the \( n \)th term of the AP \( 9, 7, 5, \dots \) is same as the \( n \)th term of \( 15, 12, 9, \dots \), find \( n \).

Solution: AP1: \( 9 + (n-1)(-2) = 11 - 2n \).
AP2: \( 15 + (n-1)(-3) = 18 - 3n \).
\( 11 - 2n = 18 - 3n \).
\( 3n - 2n = 18 - 11 \Rightarrow n = 7 \).
Answer: 7

Question 20

How many multiples of 4 lie between 10 and 250?

Solution: Multiples: \( 12, 16, \dots, 248 \).
\( 248 = 12 + (n-1)4 \)
\( 236 = 4(n-1) \Rightarrow 59 = n-1 \Rightarrow n = 60 \).
Answer: 60

This concludes the Arithmetic Progression series. Visit Omtex Classes and Omtex.co.in for other chapter resources.