Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

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 13 and common difference 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, an = a + (n –1) d

∴97 = 13 + (n –1) (4)

⇒ 4 (n –1) = 84

⇒ n – 1 = 21

⇒ n = 22

Sum of n terms of an A.P. is given by,

Thus, the required sum is 1210.

Thanks