Question

Question 13
Find the sum of first 15 multiples of 8.

Answer 1

The multiples of 8 are;
8, 16, 24, 32…
These are in an A.P having the first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
$S_{15}=?S_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{15}=\frac{15}{2}\left[2\right(8)+(15-1\left)8\right]=\frac{15}{2}[16+(14\left)\right(8\left)\right]=\frac{15}{2}(16+112)=\frac{15\left(128\right)}{2}=15\times 64=960$