question_answer

Find the sum of first 8 multiples of 3.

Answer:

First 8 multiples of 3 are 3, 6, 9............. up to 8 terms

We can observe that the above series is an AP with

\[a=3,\text{ }d=6-3=3,n=8\]

Sum of n terms of an A.P. is given by,

\[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\]

\[\therefore \]      \[{{S}_{8}}=\frac{8}{2}[2\times 3+(8-1)(3)]\]

\[=4[6+7\times 3]\]

\[=4[6+21]\]

\[=4\times 27\]

\[\Rightarrow \]   \[{{S}_{8}}=108\]