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
