Question

Find the sum of the first 15 multiples of 8.

Answer 1

The first 8 multiples of 8 are
8, 16, 24, 32, 40, 48, 56,64.......120
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
$S_{15}$ = ?
$S_n$ = $\frac{n}{2}$[2a + (n - 1)d]
$\Rightarrow$ $S_{15}$ = $\frac{15}{2}$ ​​​​[2(8) + (15 - 1)8]
$\Rightarrow$ $S_{15}$ = 7.5[6 + (14) (8)]
$\Rightarrow$ $S_{15}$ = 7.5[16 + 112]
$\Rightarrow$ $S_{15}$ = 7.5(128)
$\Rightarrow$ $S_{15}$ = 960