Question

The sum of integers divisible by $3$ from $11$ to $1000$ is

• A. $156815$
• B. $166815$
• C. $176815$
• D. $186815$

Answer 1

The correct option is B $166815$

The series of such integers are

$12,15,\mathrm{18....},999$

Here, $a=12,d=3,l=999$

$l=a+(n-1)d\Rightarrow 999=12+(n-1)3$

So, $n=330$

Sum$=\frac{n}{2}[a+l]=\frac{330}{2}[12+999]$

So, sum$=166815$