Question

Find the sum of first 8 multiples of 3.

Answer 1

First 8 multiples of 3 are 3, 6, 9 ....... upto 8 terms
We can observe that the above series is an AP with a = 3, d = 6 - 3 = 3, n = 8
Sum of n terms of an A.P. is given by,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\therefore S_8=\frac{8}{2}[2\times 3+(8-1\left)\right(3\left)\right]=4[6+7\times 3]=4[6+21]=4\times 27\Rightarrow S_8=108$