Arithmetic Progressions — Practice

Question 1

Is 2, 4, 8, 16, … an AP? Why?

Accepted Answer

Step 1 — Differences: 4−2=2, 8−4=4, not constant.

✦ Answer: No — the difference is not the same (it is a GP).

Question 2

Find the 11th term of 3, 8, 13, …

Accepted Answer

Step 1 — a = 3, d = 5.

Step 2 — a₁₁ = 3 + 10(5) = 53.

✦ Answer: 53.

Question 3

The 3rd term of an AP is 12 and the 5th is 20. Find d.

Accepted Answer

Step 1 — a₅ − a₃ = 2d ⇒ 20 − 12 = 2d ⇒ d = 4.

✦ Answer: d = 4.

Question 4

Which term of the AP 5, 11, 17, … is 95?

Accepted Answer

Step 1 — a = 5, d = 6; set 5 + (n−1)6 = 95.

Step 2 — (n−1)6 = 90 ⇒ n−1 = 15 ⇒ n = 16.

✦ Answer: the 16th term.

Question 5

Find the sum of the first 20 terms of 1, 3, 5, 7, …

Accepted Answer

Step 1 — a = 1, d = 2, n = 20.

Step 2 — S₂₀ = 20/2[2(1) + 19(2)] = 10[2 + 38] = 400.

✦ Answer: 400.

Question 6

The 4th term of an AP is 0. Prove that its 25th term is triple its 11th term.

Accepted Answer

Step 1 — a₄ = a + 3d = 0 ⇒ a = −3d.

Step 2 — a₁₁ = a + 10d = −3d + 10d = 7d; a₂₅ = a + 24d = −3d + 24d = 21d.

Step 3 — a₂₅ = 21d = 3(7d) = 3·a₁₁.

✦ Answer: proved, a₂₅ = 3·a₁₁.

Question 7

How many terms of the AP 24, 21, 18, … must be taken so that their sum is 78?

Accepted Answer

Step 1 — a = 24, d = −3; Sₙ = n/2[48 + (n−1)(−3)] = 78.

Step 2 — n(51 − 3n) = 156 ⇒ 3n² − 51n + 156 = 0 ⇒ n² − 17n + 52 = 0.

Step 3 — (n − 4)(n − 13) = 0 ⇒ n = 4 or n = 13 (both valid; sum revisits 78 as terms turn negative).

✦ Answer: n = 4 or n = 13.

Question 8

The sum of the first n terms of an AP is Sₙ = 3n² + 5n. Find its nth term and common difference.

Accepted Answer

Step 1 — aₙ = Sₙ − Sₙ₋₁ = (3n²+5n) − [3(n−1)²+5(n−1)].

Step 2 — = 3n²+5n − (3n²−6n+3+5n−5) = 6n + 2.

Step 3 — a₁ = 8, a₂ = 14 ⇒ d = 6.

✦ Answer: aₙ = 6n + 2, d = 6.

Question 9

A man repays a loan of ₹3250 by paying ₹20 in the first month and increasing the payment by ₹15 each month. How much does he pay in the 30th month?

Accepted Answer

Step 1 — a = 20, d = 15.

Step 2 — a₃₀ = 20 + 29(15) = 20 + 435 = 455.

✦ Answer: ₹455 in the 30th month.

Question 10

Find three numbers in AP whose sum is 27 and whose product is 648.

Accepted Answer

Step 1 — Take the numbers as a−d, a, a+d. Sum = 3a = 27 ⇒ a = 9.

Step 2 — Product = (9−d)(9)(9+d) = 648 ⇒ 9(81 − d²) = 648 ⇒ 81 − d² = 72 ⇒ d² = 9 ⇒ d = ±3.

Step 3 — Numbers: 6, 9, 12.

✦ Answer: 6, 9, 12.

Practice — Arithmetic Progressions

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