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

We have to first find the two-digit numbers, which when divided by 4, yield 1 as remainder.

They are: 13, 17, … 97.

As it’s seen that 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.

We know that, the n^th term of an A.P. is given by,

a_n = a + (n –1) d

So, 97 = 13 + (n –1) (4)

4 (n –1) = 84

n – 1 = 21

n = 22

Now, the sum of n terms of an A.P. is given by,

\( s_n=\dfrac{n}{2}[2a+(n-1)d]\)

\( s_{22}=\dfrac{22}{2}[2(13)+(22-1)(4)]\)

= 11[26 + 84]

= 1210

Therefore, the required sum is 1210.