A. 1012
B. 1201
C. 1212
D. 1210
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Correct Answer – C
The two-digit numbers, which when divided by `4`, yield `1` as remainder, are,
`13, 17, … 97`.
This series forms an A.P. with first term, `a = 13` and common difference, `d =4`.
Let `n` be the number of terms of the A.P.
It is known that the nth term of an A.P. is given by,` a_n = a + (n -1) d`.
`:. 97 = 13 + (n -1) (4)`
`=> 4 (n –1) = 84`
`=> n – 1 = 21`
`=> n = 22`
Now, `S_22 = n/2(2a+(n-1)d) = 22/2(2(13)+(22-1)4) = 11**110 = 1210`
So, the required sum will be `1210`.
Correct Answer – D