Question

Find the sum of first $11$ positive numbers which are multiples of $6$.

• A. $314$
• B. $396$
• C. $452$
• D. $245$

Answer 1

Answer: (B)

$396$

Clearly, the numbers are $6,12,18,...$

This is an AP with first term $a=6$, common difference $d=6$ and $n=11$.

We have to find the sum of $11$ terms of the given AP

Putting $a=6$, $d=6,n=11$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right],$ we get

$S_{11}=\frac{11}{2}[2\times 6+(11-1)\times 6]$

$\therefore S_{11}=11\times 36$

Hence,$S_{11}=396$

Option B is correct.