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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