Sum of all two digit numbers which when divided by 4 yield unity as remainder is

(b) 1210

The given series is 13, 17, 21....97.

$a_{1}=13, a_{2}=17, a_{n}=97$

$d=a_{2}-a_{1}=7-3=4$

$a_{n}=97$

$\Rightarrow a+(n-1) d=97$

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

$\Rightarrow n=22$

Sum of the above series:

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

$=11\{26+84\}$

$=1210$