Find the sum of all two digit numbers which when divided by $4,$ yields $1$ as remainder.
Find the sum of all two digit numbers which when divided by $4,$ yields $1$ as remainder.
The two-digit numbers, which when divided by $4,$ yield $1$ as remainder, are $13,17, \ldots 97$
This series forms an $A.P.$ with first term $13$ and common difference $4$
Let n be the number of terms of the $A.P.$
It is known that the $n^{th}$ term of an $A.P.$ is given by, $a_{n}=a+(n-1) d$
$\therefore 97=13+(n-1)(4)$
$\Rightarrow 4(n-1)=84$
$\Rightarrow n-1=21$
$\Rightarrow n=22$
Sum of n terms of an $A.P.$ is given by
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\therefore S_{22}=\frac{22}{2}[2(13)+(22-1)(4)]$
$=11[26+84]$
$=1210$
Thus, the required sum is $1210 .$
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