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

Find the sum of first 15 multiples of 8.

Solution:
The first 15 multiples of 8 are 8, 16, 24, 32,…
This is an AP in which a = 8, d = (16 – 8) = 8 and n = 15.
Thus, we have:
$l=a+(n-1) d$
$=8+(15-1) 8$
$=120$
$\therefore$ Required sum $=\frac{n}{2}(a+l)$
$=\frac{15}{2}[8+120]=15 \times 64=960$
Hence, the required sum is 960.
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