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.

Given:

4th term = a₄ = 9

Sum of 6th and 13th terms = a₆ + a₁₃ = 40

Now,

a₄ = a + (4-1)d

9 = a + 3d

a = 9 – 3d ….(1)

And

a₆ + a₁₃ = 40

a + 5d + a + 12d = 40

2a +17d = 40

2(9 – 3d) + 17d = 40 (using (1))

d = 2

From (1): a = 9 – 6 = 3

Required AP = 3,5,7,9,…..