Question

The sum of the first 7 terms of an AP is 182. If its 4^th and 17^th  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 = 8 

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