Question

Find the sum of all multiples of $9$ lying between $300$ and $700$.

• A. $21978$
• B. $22978$
• C. $21970$
• D. $21960$

Answer 1

The correct option is A $21978$

Clearly, the numbers between $300$ and $700$ which are multiples of $9$ are $306,315,324,...,693$.

This is an AP with first term $a=306$, common difference $d=9$ and last term $l=693$.
Let there be $n$ terms in this AP. Then,
$a_n=693\Rightarrow a+(n-1)d$

$693\Rightarrow 306+(n-1)\times 9$

$\therefore n=44$
$\therefore$ Required sum $=S_n=\frac{n}{2}[a+l]=\frac{44}{2}[306+693]=21978$