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 \(d = 8\)
\(S_{15} =?\)
\(S_n = \frac n2 [2a +(n-1)d]\)

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

\(S_{15} = \frac{15}{2} [16 +14(8)]\)

\(S_{15}= \frac {15}{2} [16 +112]\)

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