Find the sum of first 15 multiples of 8.

The multiples of 8 are 8, 16, 24, 32…

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore, a = 8
d = 8
We have to find \(S_{15}\)

By the formula of sum of \(n^{th}\) term, we know,

\(S_n = {{n} \over {2}} [2a+(n-1)d]\)
\(S_{15} = {{15} \over {2}} [2(8) + (15-1)8] \)
\(= {{15} \over {2}}[6 +(14)(8)] \)
\( = {{15} \over {2}}[16 +112]\)
\(= {{15(128)} \over {2}}\)
\(= 15 × 64\)
\(= 960\)