Question

Find the sum of all the three digit natural numbers which on division by $7$ leaves remainder $3.$

Answer 1

As per the question,

All these numbers are $101,108,115,...,997$

Here the last term is $l=997$

We know that, In A.P,

$a_n=a+(n-1)d$
$\Rightarrow 997=101+(n-1)7$

$\Rightarrow 997=101+7n-7$

$\Rightarrow 997-101+7=7n$

$\Rightarrow 903=7n$

$\Rightarrow n=\frac{903}{7}$
$\Rightarrow$ $n=129$

Now, Sum of $n$ terms in A.P,

$S_n=\frac{n}{2}(a+l)$

$\Rightarrow S_n=\frac{129}{2}[101+997].$

$S_n=\frac{129\times 1098}{2}$

$S_n=129\times 549$

$S_n=70821$

Hence, required sum is $70821$