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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.
Solution
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, Sn = n2 + 6n
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