27 Arithmetic Progression Practice Questions with Solutions

27 Arithmetic Progression Questions

Set 4: Sharpen your skills with Omtex Classes.

Part 1: Basic Calculations

Question 1

Find the next term of the AP: \( \sqrt{7}, \sqrt{28}, \sqrt{63}, \dots \)

Solution: Simplify terms: \( \sqrt{7} \), \( \sqrt{4 \times 7} = 2\sqrt{7} \), \( \sqrt{9 \times 7} = 3\sqrt{7} \).
This is an AP with \( d = \sqrt{7} \).
Next term: \( 4\sqrt{7} = \sqrt{16 \times 7} = \sqrt{112} \).
Answer: \( \sqrt{112} \)

Question 2

Is 302 a term of the AP \( 3, 8, 13, \dots \)?

Solution: \( a=3, d=5 \).
\( 302 = 3 + (n-1)5 \Rightarrow 299 = 5(n-1) \).
\( n-1 = 59.8 \). Since \( n \) is not an integer, it is not a term.
Answer: No

Question 3

If \( a_n = 3n^2 - 4n \), find the 5th term.

Solution: Note: This formula usually represents \( S_n \) in quadratics, but if given as \( a_n \), simply substitute.
\( a_5 = 3(5)^2 - 4(5) = 3(25) - 20 = 75 - 20 = 55 \).
Answer: 55

Question 4

How many multiples of 6 lie between 1 and 100?

Solution: AP: \( 6, 12, \dots, 96 \).
\( 96 = 6 + (n-1)6 \Rightarrow 90 = 6(n-1) \Rightarrow 15 = n-1 \Rightarrow n=16 \).
Answer: 16

Question 5

Find the common difference of the AP whose general term is \( a_n = 3n + 7 \).

Solution: \( a_1 = 10, a_2 = 13 \).
\( d = 13 - 10 = 3 \).
(Shortcut: The coefficient of \( n \) is the common difference).
Answer: 3

Question 6

Find the sum of the first 20 even natural numbers.

Solution: AP: \( 2, 4, 6, \dots \). \( a=2, d=2, n=20 \).
\( S_{20} = \frac{20}{2}[2(2) + 19(2)] = 10(4 + 38) = 10(42) = 420 \).
Or use formula \( n(n+1) = 20 \times 21 = 420 \).
Answer: 420

Question 7

If the 1st term is 7 and the 13th term is 35, find the common difference.

Solution: \( a_{13} = a + 12d \Rightarrow 35 = 7 + 12d \).
\( 28 = 12d \Rightarrow d = 28/12 = 7/3 \).
Answer: 7/3

Question 8

Check if the list of numbers \( 1^2, 3^2, 5^2, 7^2, \dots \) forms an AP.

Solution: Terms: \( 1, 9, 25, 49 \).
\( 9-1 = 8 \). \( 25-9 = 16 \).
Differences are not constant.
Answer: No

Question 9

If 5 times the 5th term of an AP is equal to 8 times its 8th term, find the 13th term.

Solution: \( 5(a+4d) = 8(a+7d) \)
\( 5a + 20d = 8a + 56d \)
\( -3a = 36d \Rightarrow a = -12d \Rightarrow a + 12d = 0 \).
\( a_{13} = a + 12d = 0 \).
Answer: 0

Question 10

Find the sum of \( 2+4+6+\dots+200 \).

Solution: \( a=2, l=200, n=100 \).
\( S_{100} = \frac{100}{2}(2+200) = 50(202) = 10100 \).
Answer: 10100

Part 2: Intermediate Questions

Question 11

How many two-digit numbers are divisible by 6?

Solution: First: 12, Last: 96. \( d=6 \).
\( 96 = 12 + (n-1)6 \Rightarrow 84 = 6(n-1) \Rightarrow 14 = n-1 \Rightarrow n=15 \).
Answer: 15

Question 12

Which term of the AP \( 53, 48, 43, \dots \) is the first negative term?

Solution: \( a=53, d=-5 \).
\( 53 + (n-1)(-5) < 0 \Rightarrow 53 - 5n + 5 < 0 \)
\( 58 < 5n \Rightarrow n > 11.6 \).
First integer is 12.
Answer: 12th term

Question 13

Find the sum of all three-digit numbers divisible by 9.

Solution: AP: \( 108, 117, \dots, 999 \).
\( 999 = 108 + (n-1)9 \Rightarrow 891 = 9(n-1) \Rightarrow 99 = n-1 \Rightarrow n=100 \).
\( S_{100} = \frac{100}{2}(108+999) = 50(1107) = 55350 \).
Answer: 55350

Question 14

The angles of a triangle are in AP. The largest angle is 90°. Find the other angles.

Solution: Angles: \( a, a+d, a+2d \). \( a+2d = 90 \).
Sum: \( 3a+3d = 180 \Rightarrow a+d = 60 \).
\( a+d \) is the middle angle, so \( 60^\circ \).
AP: \( 30^\circ, 60^\circ, 90^\circ \).
Answer: 30°, 60°, 90°

Question 15

Insert three arithmetic means between 3 and 19.

Solution: AP: \( 3, A_1, A_2, A_3, 19 \). Total 5 terms.
\( 19 = 3 + 4d \Rightarrow 16 = 4d \Rightarrow d = 4 \).
Terms: \( 3+4=7 \), \( 7+4=11 \), \( 11+4=15 \).
Answer: 7, 11, 15

Question 16

For what value of \( k \) are \( k, 2k-1, 2k+1 \) in AP?

Solution: \( 2(2k-1) = k + (2k+1) \)
\( 4k - 2 = 3k + 1 \)
\( k = 3 \).
Answer: 3

Question 17

Find the sum of the first \( n \) odd natural numbers.

Solution: \( 1, 3, 5, \dots \). \( a=1, d=2 \).
\( S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2 \).
Answer: \( n^2 \)

Question 18

If the \( p \)th term of an AP is \( q \) and the \( q \)th term is \( p \), show that the \( n \)th term is \( p+q-n \).

Solution: \( a+(p-1)d = q \) and \( a+(q-1)d = p \).
Subtracting: \( d(p-q) = q-p = -(p-q) \Rightarrow d = -1 \).
Substitute \( d \): \( a - p + 1 = q \Rightarrow a = p+q-1 \).
\( a_n = p+q-1 + (n-1)(-1) = p+q-1 - n + 1 = p+q-n \).
Answer: Proved

Part 3: Word Problems & Proofs

Question 19

A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2.5 m apart, what is the length of the wood required for the rungs?

Solution: Total height = 250 cm. Gap = 25 cm.
Number of rungs = \( \frac{250}{25} + 1 = 11 \).
AP: \( a=45, l=25, n=11 \).
\( S_{11} = \frac{11}{2}(45+25) = \frac{11}{2}(70) = 11 \times 35 = 385 \).
Answer: 385 cm

Question 20

If \( a, b, c \) are in AP, prove that \( \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab} \) are also in AP.

Solution: If they are in AP, then \( \frac{1}{ca} - \frac{1}{bc} = \frac{1}{ab} - \frac{1}{ca} \).
Multiply entire equation by \( abc \):
\( b - a = c - b \Rightarrow 2b = a+c \).
Since \( a, b, c \) are in AP, \( 2b=a+c \) is true.
Answer: Proved

Question 21

The sum of the first \( n \) terms of an AP is given by \( S_n = \frac{5n^2}{2} + \frac{3n}{2} \). Find the 20th term.

Solution: \( a_{20} = S_{20} - S_{19} \).
\( S_{20} = \frac{5(400)}{2} + \frac{60}{2} = 1000 + 30 = 1030 \).
\( S_{19} = \frac{5(361)}{2} + \frac{57}{2} = \frac{1805+57}{2} = \frac{1862}{2} = 931 \).
\( a_{20} = 1030 - 931 = 99 \).
Answer: 99

Question 22

If the ratio of the sum of the first \( n \) terms of two APs is \( (7n+1):(4n+27) \), find the ratio of their 9th terms.

Solution: Ratio of \( m \)th terms corresponds to replacing \( n \) with \( 2m-1 \) in sum ratio.
Here \( m=9 \), so replace \( n \) with \( 2(9)-1 = 17 \).
Ratio = \( \frac{7(17)+1}{4(17)+27} = \frac{119+1}{68+27} = \frac{120}{95} \).
Simplify: \( 24:19 \).
Answer: 24:19

Question 23

If \( a, b, c \) are in AP, prove that \( (a-c)^2 = 4(b^2 - ac) \).

Solution: Since AP, \( b = \frac{a+c}{2} \).
RHS: \( 4[(\frac{a+c}{2})^2 - ac] = 4[\frac{a^2+c^2+2ac}{4} - ac] \)
\( = a^2 + c^2 + 2ac - 4ac = a^2 + c^2 - 2ac = (a-c)^2 \).
LHS = RHS.
Answer: Proved

Question 24

The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term.

Solution: \( S_7 = 63 \Rightarrow \frac{7}{2}(2a+6d)=63 \Rightarrow a+3d=9 \).
\( S_{14} = 63 + 161 = 224 \Rightarrow \frac{14}{2}(2a+13d)=224 \Rightarrow 7(2a+13d)=224 \Rightarrow 2a+13d=32 \).
Solve: \( 2(9-3d)+13d=32 \Rightarrow 18-6d+13d=32 \Rightarrow 7d=14 \Rightarrow d=2 \).
\( a = 9 - 6 = 3 \).
\( a_{28} = 3 + 27(2) = 57 \).
Answer: 57

Question 25

Solve for \( x \): \( 1 + 6 + 11 + 16 + \dots + x = 148 \).

Solution: \( a=1, d=5, S_n=148 \).
\( 148 = \frac{n}{2}[2 + (n-1)5] \Rightarrow 296 = 2n + 5n^2 - 5n = 5n^2 - 3n \).
\( 5n^2 - 3n - 296 = 0 \). Using quadratic formula, \( n=8 \).
\( x = a_8 = 1 + 7(5) = 36 \).
Answer: 36

Question 26

The sum of three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers.

Solution: Terms: \( a-d, a, a+d \). Sum \( 3a=12 \Rightarrow a=4 \).
\( (4-d)^3 + 4^3 + (4+d)^3 = 288 \).
\( 64 + (64 - 48d + 12d^2 - d^3) + (64 + 48d + 12d^2 + d^3) = 288 \).
\( 192 + 24d^2 = 288 \Rightarrow 24d^2 = 96 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Numbers: \( 2, 4, 6 \).
Answer: 2, 4, 6

Question 27

Split 24 into three parts such that they are in AP and their product is 440.

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

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