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

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

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

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

It is known that the $n^{th}$ term of an $A.P.$ is given by, $a_{n}=a+(n-1) d$

$\therefore 97=13+(n-1)(4)$

$\Rightarrow 4(n-1)=84$

$\Rightarrow n-1=21$

$\Rightarrow n=22$

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

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore S_{22}=\frac{22}{2}[2(13)+(22-1)(4)]$

$=11[26+84]$

$=1210$

Thus, the required sum is $1210 .$