Question

Find the sum of the first 15 multiples of 8.

Answer 1

The multiples of 8 are
$8, 16, 24, 32….$
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, $a = 8$ and $d = 8$
$S_{15} =?$
$S_n = \frac n2 [2a +(n-1)d]$

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

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

$S_{15}= \frac {15}{2} [16 +112]$

$S_{15}= \frac {15 \times (128)}{2}$
$S_{15}= 15 \times 64$
$S_{15}= 960$