Find the sum of all two digit numbers which when divided by 4,

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

$\therefore a=13, d=4, a_{n}=97$

$\therefore a_{n}=a+(n-1) d$

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

$\Rightarrow 84=4 n-4$

$\Rightarrow 88=4 n$

$\Rightarrow 22=n \ldots(1)$

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

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

$\Rightarrow S_{22}=11[110]=1210$