Find the sum of the first 15 multiples of 8.

Solution:

The multiples of 8 are 8, 16, 24, 3...

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

$S_{15}=?$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{15}{2}[2(8)+(15-1) 8]$

$=\frac{15}{2}[16+14(8)]$

$=\frac{15}{2}(16+112)$

$=\frac{15(128)}{2}=15 \times 64$

$=960$