Question

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

Answer 1

The first two digit number which when divided by $4$ leaves remainder $1$ is $43+1=13$ and last is $424+1=97$; form $(4k+1)$.
Thus we have to find the sum of $13+17+21+$______$+97$,
which is an A.P. $\therefore 97=13+(n-1)\cdot 4$
$\therefore n=22$
$\therefore S=\left(n/2\right)[a+l]=11(13+97)$
$=11\times 110=1210$.