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Q.
The sum of all two digit numbers which when divided by $4$, yield unity as remainder, is
Sequences and Series
Solution:
The first two digit number which when divided by $4$ leaves remainder $1$ is $4 \cdot 3+1=13$ and last is $4 \cdot 24 +1=97 .$
Thus, we have to find the sum
$13+17+21+\ldots+97$
which is an A.P.
$ \therefore 97 =13+(n-1) \cdot 4 $
$\Rightarrow n=22$
and $ S_{n}=[a+l]=11 \times[13+97] $
$=11 \times 110=1210$