Question

How many terms of the series 54, 51, 48, .... be taken so that their sum is 513 ? Explain the answer.

Answer 1

Since,

a = 54, d = – 3 and $S_n$ = 513

($\frac{n}{2}$)[2a + (n – 1) d] = 513

($\frac{n}{2}$)[108 + (n – 1) $\times$ – 3] = 513

$n^2$ – 37n + 342 = 0

(n – 18) (n ­– 19) = 0

n = 18 or 19

Here, the common difference is negative.

So, 19th term is $a_{19}$ = 54 + (19 – 1) $\times$ – 3 = 0.

Thus, the sum of 18 terms as well as that of 19 terms is 513.