The sum of all two digit numbers which when divided by 4, leave 1 as remainder is :

Correct option is D

Given:

Need the sum of all two-digit numbers that leave remainder 1 when divided by 4.

Formula Used:

Sum of an A.P. is given by

​$$S_n = \frac{n}{2}(a + l)$$​

where n is the number of terms, a is the first term, and l is the last term.

Solution:

First two-digit number leaving remainder 1 when divided by 4: 13.

Last two-digit number leaving remainder 1 when divided by 4: 97.

Now, common difference d = 4.

Finding number of terms:

​$$n = \frac{l - a}{d} + 1 = \frac{97 - 13}{4} + 1 = \frac{84}{4} + 1 = 21 + 1 = 22$$​​

Now, sum:

​$$S_{22} = \frac{22}{2}(13 + 97) = 11 \times 110 = 1210$$​​

Therefore, the required sum is 1210.