Question

The sum of all three-digit natural numbers which are divisible by $7$, is

• A. $25501$
• B. $52005$
• C. $70336$
• D. $84321$

Answer 1

The correct option is C $70336$

The smallest and the largest three digit-number, which are divisible by $7$ are $105$ and $994$, respectively.
So, the sequence of three-digit numbers which are divisible by $7$ is $105,112,119,\dots ,994.$

Clearly, it is an A.P. with first term $a=105$ and common difference $d=7.$
Let there be $n$ terms in this sequence.
Then, $a_n=994$
$\Rightarrow a+(n-1)d=994\Rightarrow 105+(n-1)7=994\Rightarrow n=128$

Now, required sum
$=\frac{n}{2}[a+l]=\frac{128}{2}[105+994]=70336$