Question

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

Answer 1

Required sum

13 + 17 + 21 + 25 + ... + 97.

This is an AP in which a= 13, d = (17 - 13) = 4 and l= 97.

Let the number of terms be n. Then,

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

$\Rightarrow 13+(n-1)\times 4=97\Rightarrow n=22$

$\therefore$ required sum $=\frac{n}{2}(a+1)=\frac{22}{2}\times (13+97)=(11\times 110)=1210$

Hence, the required sum is 1210.