Question

Find the sum of first $32$ multiples of $4$.

• A. $2,112$
• B. $2,712$
• C. $2,110$
• D. $2,111$

Answer 1

The correct option is A $2,112$

The multiples of $4$ are, $4,8,12,16,$ ............
Therefore, $a=4,d=4,n=32$
We have $S_{32}=\frac{n}{2}[2a+(n-1\left)d\right]$
$=\frac{32}{2}[2\times 4+(32-1\left)4\right]$
$=16[8+31\times 4]$
$=16[8+124]$
Therefore, $S_{20}=2,112$