Q20 of 805 Page 1

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

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

We know, nth term of an AP


an = a + (n – 1)d and


Sum of ‘n’ terms of an AP


∴ T4 = a + 3d = 9


⇒ a = 9 – 3d [1]


and S6 + S13 = 40



⇒ 19a + 93d = 40


⇒ 19(9 – 3d) + 93d = 40 [From [1]]


⇒ 171 + 36d = 40




Then, T1 = a =


T2 = a + d =


T3 = a + 2d


T4 = a + 3d


So, A.P. is

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