Q24 of 39 Page 1

If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Given,
S4 = 40

S14 = 280

We know that,



⇒ 2(2a + 3d) = 40

⇒ 4a + 12d = 40

⇒ 2a + 3d = 20 …(1)


⇒ 7(2a + 13d) = 280

⇒ 14a + 91 = 280

⇒ 2a + 13d = 40 …(2)


Subtract eq 1 from eq 2 to get,

⇒ 2a + 13d - (2a + 3d) = 40 - 20

⇒ 2a + 13d - 2a - 3d = 40 - 20

⇒  10d = 20

⇒ d = 2

Put the value of d in (1) to get,

⇒ 2a + 3(2) = 20

⇒ 2a + 6 = 20

⇒ 2a = 20 – 6

⇒ 2a = 14

⇒ a = 7




= n2 + 6n

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