Question

Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Answer 1

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

a₄ = 9 

⇒ a + (4-1) d = 9                        [ a_n = a + (n-1) d]

⇒ a +3d = 9         ....................(1)

Now,

a₆ +a₁₃ = 40           (Given) 

⇒ (a +5d ) + (a +12d) = 40

⇒ 2a + 17d = 40            ...............(2)

From (1) and (2), we get

2(9-3d ) +17d = 40

⇒ 18-6d + 17d = 40

⇒ 11d = 40 - 18 =22

⇒ d =2

Putting d = 2 in (1), we get

a +3 × 2 = 9

⇒ a = 9-6=3

Hence, the AP is 3, 5, 7, 9, 11,…….