Question

The sum of all natural numbers which are divisible by 3 and less than 100 is

• A. 1863
• B. 1368
• C. 1683
• D. 1763

Answer 1

The correct option is C

1683

The numbers are $3,6,\mathrm{9.......99}$.
Let $n$ be the total number of terms on the given A.P.
$\Rightarrow a+(n-1)d=99$
$\Rightarrow 3+(n-1)3=99$
$\Rightarrow 3(n-1)=96$
$\Rightarrow n-1=32$
$\Rightarrow n=33$
Now, sum till $n$ terms is given by
$S=\frac{n}{2}(a+l)$
$=\frac{33}{2}\times (99+3)=1683$