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Question

The sum of all two digit numbers which when divided by 4, yied unity as remainder, is

A
1012
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B
1201
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C
1212
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D
1210
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Solution

The correct option is D 1210
According to given conditions , such numbers are
13,17,21,........................97
which is an AP with a=13,d=4
so, that a0=13,a1=17......an=a+(n1)d

97=13+(n1)4n=22

Now, Sum of AP
=n2(2a+(n1)d)=222(213+(221)4)=11110=1210









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