Question

Find the sum of all two digit numbers which when divided by $4$, yields $1$ as remainder

Answer 1

The two digit numbers, which when divided by $4$, yield $1$ as remainder are, $13,17,...,97$.

This series forms an A.P. with first term $13$ and common difference is $4$.

Let $n$ be the number of terms of A.P.

It is known that the $n^{th}$ term of an A.P. is given by,

$\therefore a_n=a+(n-1)d\therefore 97=13+(n-1)\left(4\right)\Rightarrow 4(n-1)=84\Rightarrow n-1=21\Rightarrow n=22$

Sum of $n$ terms of an A.P. is given by

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{22}=\frac{22}{2}\left[2\right(13)+(22-1\left)\right(4\left)\right]=11[26+84]=1210$

Thus the required sum is $1210.$