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, … 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 $n^{\text {th }}$ 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,

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

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

$=11[26+84]$

$=1210$

Thus, the required sum is 1210.