Question

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

Answer 1

Given that S4 = 40 and S14 = 280'

`"S"_"n" = "n"/2[2"a" + ("n-1)d"]`

`"S"_4 = 4/2[2"a" + (4-1)"d"] = 40`

`=> 2"a" + 3"d" = 20` .......(i)

`"S"_14 = 14/2 [2"a" + (14 -1)"d"] = 280`

`=> 2"a" + 13"d" = 40` ...(ii)

(ii) - (i)

10d = 20 ⇒ d = 2

Sunstituting the value of d in (i) we get

2a + 6 = 20 ⇒ a = 7

Sum of first n terms,

`"S"_"n" = "n"/2[2"a" + ("n-1)d"]`

= `"n"/2 [14 + ("n"-1) 2]`

= n ( 7 + n - 1)

= n (n + 6)

= n² + 6n

Therefore, S_n = n² + 6n