MCQ
The sum of all two digit numbers which, when divided by $4$, yield unity as a remainder is
- A$1190$
- B$1197$
- ✓$1210$
- DNone of these
This is an $AP$ with first term $13$ and common difference $4$.
Let the number of terms be $n$.
Then $97 = 13 + (n - 1)4$
$ \Rightarrow $ $4n = 88$
$ \Rightarrow $ $n = 22$
Therefore the sum of the numbers
$ = \frac{{22}}{2}[13 + 97] = 11(110) = 1210$.
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