Question

The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1 : 5, find the AP.

Answer 1

Let a be the first term and d be the common difference of the AP.

∴ S₇ = 182 

⇒ `7/2 (2a + 6d) = 182`   ...`(S_n = n/2 [2a + (n - 1)d]}`

⇒ a + 3d = 26   ...(1)

Also,

a₄ : a₁₇ = 1 : 5   ...(Given)

⇒ `(a + 3d)/(a + 16d) = 1/5`   ...[a_n = a + (n – 1)d]

⇒ 5a + 15d = a + 16d

⇒ d = 4a   ...(2)

Solving (1) and (2), we get

a + 3 × 4a = 26

⇒ 13a = 26

⇒ a = 2

Putting a = 2 in (2), we get

d = 4 × 2 

d = 8 

Hence, the required AP is 2, 10, 18, 26,.........