28 Important Arithmetic Progression Questions with Solutions

28 Arithmetic Progression Questions

Curated for excellence by Omtex Classes.

Part 1: Calculations and Terms

Question 1

Find the common difference and the next term of the AP: \( 0.6, 1.7, 2.8, \dots \)

Solution: \( a_1 = 0.6, a_2 = 1.7 \).
\( d = 1.7 - 0.6 = 1.1 \).
Next term \( = 2.8 + 1.1 = 3.9 \).
Answer: d=1.1, Next term=3.9

Question 2

In an AP, if \( d = -4, n = 7, a_n = 4 \), then find \( a \).

Solution: \( a_n = a + (n-1)d \).
\( 4 = a + (7-1)(-4) \)
\( 4 = a - 24 \Rightarrow a = 28 \).
Answer: 28

Question 3

Find the missing terms in the boxes: \( \Box, 38, \Box, \Box, \Box, -22 \).

Solution: \( a_2 = 38 \Rightarrow a+d = 38 \) (i)
\( a_6 = -22 \Rightarrow a+5d = -22 \) (ii)
Subtract (i) from (ii): \( 4d = -60 \Rightarrow d = -15 \).
From (i): \( a - 15 = 38 \Rightarrow a = 53 \).
Terms: 53, 38, 23, 8, -7, -22.
Answer: 53, 23, 8, -7

Question 4

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution: Let APs be \( a, a+d, \dots \) and \( b, b+d, \dots \).
Diff of 100th terms: \( (a+99d) - (b+99d) = a - b = 100 \).
Diff of 1000th terms: \( (a+999d) - (b+999d) = a - b \).
Since \( a-b = 100 \), the difference remains the same.
Answer: 100

Question 5

How many multiples of 4 lie between 10 and 250?

Solution: First multiple > 10 is 12. Last multiple < 250 is 248.
\( 248 = 12 + (n-1)4 \)
\( 236 = 4(n-1) \Rightarrow 59 = n - 1 \Rightarrow n = 60 \).
Answer: 60

Question 6

Find the sum of the first 100 natural numbers.

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

Question 7

If the sum of first \( n \) terms of an AP is \( 4n - n^2 \), what is the first term? What is the sum of first two terms? What is the 2nd term?

Solution: \( S_n = 4n - n^2 \).
\( S_1 = 4(1) - 1^2 = 3 \). (First term \( a_1 = 3 \)).
\( S_2 = 4(2) - 2^2 = 8 - 4 = 4 \).
\( a_2 = S_2 - S_1 = 4 - 3 = 1 \).
Answer: First term=3, Sum(2)=4, 2nd term=1

Question 8

Determine the AP whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.

Solution: \( a_7 - a_5 = 12 \Rightarrow (a+6d) - (a+4d) = 12 \Rightarrow 2d = 12 \Rightarrow d = 6 \).
\( a_3 = 16 \Rightarrow a + 2d = 16 \Rightarrow a + 12 = 16 \Rightarrow a = 4 \).
AP: \( 4, 10, 16, \dots \).
Answer: 4, 10, 16...

Question 9

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

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

Question 10

Check whether -150 is a term of the AP: \( 11, 8, 5, 2, \dots \)

Solution: \( a=11, d=-3 \).
\( -150 = 11 + (n-1)(-3) \Rightarrow -161 = -3(n-1) \).
\( n-1 = 161/3 = 53.66 \).
Since \( n \) is not a whole number, it is not a term.
Answer: No

Part 2: Sums and Word Problems

Question 11

Find the sum of the first 15 multiples of 8.

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

Question 12

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

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

Question 13

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... What is the total length of such a spiral made up of 13 consecutive semicircles? (\( \pi = 22/7 \))

Solution: Lengths form AP: \( \pi(0.5), \pi(1.0), \dots \)
\( a=0.5\pi, d=0.5\pi, n=13 \).
Total Length \( = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[7\pi] \).
\( = \frac{91}{2} \times \frac{22}{7} = 13 \times 11 = 143 \) cm.
Answer: 143 cm

Question 14

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

Solution: \( S_n=200, a=20, d=-1 \).
\( 200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = 41n - n^2 \).
\( n^2 - 41n + 400 = 0 \). Factors: 16, 25.
If \( n=25, a_{25} = 20-24 = -4 \) (Impossible).
So \( n=16 \). Logs in top row \( a_{16} = 20 - 15 = 5 \).
Answer: 16 rows, 5 logs

Question 15

Find the sum of odd numbers between 0 and 50.

Solution: Terms: \( 1, 3, 5, \dots, 49 \). \( n=25 \).
\( S_{25} = \frac{25}{2}(1+49) = \frac{25}{2}(50) = 625 \).
Answer: 625

Question 16

A contract on construction specifies a penalty for delay: Rs 200 for the first day, Rs 250 for the second, Rs 300 for the third, etc. How much money does the contractor have to pay as penalty for 30 days delay?

Solution: \( a=200, d=50, n=30 \).
\( S_{30} = \frac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] \).
\( = 15(1850) = 27750 \).
Answer: Rs 27,750

Part 3: Advanced & HOTS

Question 17

If the numbers \( x-2, 4x-1, \) and \( 5x+2 \) are in AP, find the value of \( x \).

Solution: \( 2(4x-1) = (x-2) + (5x+2) \)
\( 8x - 2 = 6x \)
\( 2x = 2 \Rightarrow x = 1 \).
Answer: 1

Question 18

Find the middle term of the AP: \( 213, 205, 197, \dots, 37 \).

Solution: \( a=213, d=-8, a_n=37 \).
\( 37 = 213 + (n-1)(-8) \Rightarrow -176 = -8(n-1) \Rightarrow 22 = n-1 \Rightarrow n=23 \).
Middle term = \( \frac{23+1}{2} = 12 \)th term.
\( a_{12} = 213 + 11(-8) = 213 - 88 = 125 \).
Answer: 125

Question 19

Which term of the sequence \( 20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \dots \) is the first negative term?

Solution: \( a=20, d = -0.75 \).
\( 20 + (n-1)(-0.75) < 0 \)
\( 20 < 0.75(n-1) \Rightarrow \frac{20}{0.75} < n-1 \).
\( 26.66 < n-1 \Rightarrow n > 27.66 \).
First integer is 28.
Answer: 28th term

Question 20

If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that the 18th term is zero.

Solution: \( 7(a+6d) = 11(a+10d) \)
\( 7a + 42d = 11a + 110d \)
\( -4a = 68d \Rightarrow a = -17d \).
\( a_{18} = a + 17d = -17d + 17d = 0 \).
Answer: Proved

Question 21

Find the sum of all three-digit natural numbers which are divisible by 7.

Solution: First: 105, Last: 994.
\( 994 = 105 + (n-1)7 \Rightarrow 889 = 7(n-1) \Rightarrow 127 = n-1 \Rightarrow n=128 \).
\( S_{128} = \frac{128}{2}(105+994) = 64(1099) = 70336 \).
Answer: 70336

Question 22

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

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

Question 23

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

Solution: \( a=1, d=3 \).
\( 287 = \frac{n}{2}[2 + (n-1)3] \Rightarrow 574 = n(3n-1) \).
\( 3n^2 - n - 574 = 0 \). Using quadratic formula, \( n=14 \).
\( x = a_{14} = 1 + 13(3) = 40 \).
Answer: x = 40

Question 24

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of \( x \) such that the sum of the numbers of the houses preceding the house numbered \( x \) is equal to the sum of the numbers of the houses following it. Find \( x \).

Solution: Sum 1 to \( x-1 \) = Sum \( x+1 \) to 49.
\( \frac{x-1}{2}(1 + x-1) = S_{49} - S_x \).
\( \frac{x(x-1)}{2} = \frac{49 \times 50}{2} - \frac{x(x+1)}{2} \).
\( x^2 - x = 2450 - (x^2 + x) \)
\( 2x^2 = 2450 \Rightarrow x^2 = 1225 \Rightarrow x = 35 \).
Answer: 35

Question 25

If \( m \) times the \( m \)th term of an AP is equal to \( n \) times its \( n \)th term, then show that the \( (m+n) \)th term is 0.

Solution: \( m[a+(m-1)d] = n[a+(n-1)d] \).
\( am + m^2d - md = an + n^2d - nd \).
\( a(m-n) + d(m^2-n^2) - d(m-n) = 0 \).
Divide by \( m-n \): \( a + d(m+n) - d = 0 \).
\( a + (m+n-1)d = 0 \). This is \( a_{m+n} \).
Answer: Proved

Question 26

Calculate the common difference of an AP where the first term is 100, and the sum of the first 6 terms is 5 times the sum of the next 6 terms.

Solution: \( S_6 = 5(S_{12} - S_6) \Rightarrow 6S_6 = 5S_{12} \).
\( 6[\frac{6}{2}(2a+5d)] = 5[\frac{12}{2}(2a+11d)] \).
\( 18(200+5d) = 30(200+11d) \).
Dividing by 6: \( 3(200+5d) = 5(200+11d) \).
\( 600 + 15d = 1000 + 55d \).
\( -400 = 40d \Rightarrow d = -10 \).
Answer: -10

Question 27

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7:15. Find the numbers.

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

Question 28

Show that the sequence \( \log a, \log(ab), \log(ab^2), \dots \) is an AP. Find the \( n \)th term.

Solution: \( a_1 = \log a \).
\( a_2 = \log a + \log b \).
\( a_3 = \log a + 2\log b \).
Common difference \( d = \log b \).
\( a_n = \log a + (n-1)\log b = \log(ab^{n-1}) \).
Answer: Yes, \( \log(ab^{n-1}) \)

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