Question

What is the sum of first five positive integers that are divisible by $6$.

Answer 1

Hint: formula used $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$, where $S_n$ is the sum, $a$ is the initial term, $n$ is the number of terms, and $d$ is the common difference.

Concept Used: sum of an A.P.

Prerequisites: an A.P.

Calculating the sum of the required first five terms,

Here $a=6,n=5,$ and $d=6$ because the A.P. is $6,12,18,24,$ and $30$.

$\begin{array}{rcl}{S}_n& =& \frac{n}{2}\left(2a+\left(n-1\right)d\right)\\ & =& \frac{5}{2}\left(2\times 6+\left(5-1\right)6\right)\\ & =& 2.5\left(36\right)\\ & =& 90\end{array}$

Hence, the sum of the term is $90.$