Q16 of 55 Page 1

Find an A.P. whose fourth term is 9 and the sum of its sixth term and the thirteenth term is 40.

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


Given: a4 = 9


a6 + a13 = 40


We first find the common difference.


Now, Consider a4 = 9


⇒ a + (4 – 1)d = 9


⇒ a + 3d = 9 ……………………….(1)


Consider a6 + a13 = 40


⇒ a + (6 – 1)d + a + (13 – 1)d = 40


⇒ 2a + 17d = 40 ………………..(2)


Subtracting twice of equation (1) from equation (2), we get,


11d = 22


⇒ d = 2


∴ Common difference = d = 2


Put the value of d in equation (), we get


a = 9 – 3d


= 9 – 6


= 3


∴ a = 3


The AP will be a, a + d, a + 2d, a + 3d,…


Thus the AP will be 3, 5, 7, 9,…


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