Question

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

Answer 1

Two digit numbers are:

10, 11, 12, .... , 97.

Here, a = 13, d = 17 - 13 = 4 and $a_n=97$

$\because a_n=a+(n-1)d$

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

$\Rightarrow n-1=\frac{97-13}{4}$

$\Rightarrow n-1=\frac{84}{4}=21$

$\Rightarrow n=21+1=22$

Now, $S_n=\frac{n}{2}(a+a_n)$

$\Rightarrow S_{22}=\frac{22}{2}(13+97)$

$=11\left(110\right)=1210$.