Question

The sum of the first fifteen multiples of 8 is
(a) 840
(b) 960
(c) 872
(d) 1081

Answer 1

(b) 960

The first 15 multiples of 8 are 8,16,24, 32, ..., 120.
This is an AP in which a = 8, d = (16 - 8) = 8, l = 120 and n = 15

∴ Required sum = $\frac{n}{2}\left(a+l\right)$
= $\left(\frac{15}{2}\right)\times \left(8+120\right)=15\times 64=960$

Hence, the required sum is 960.