# Find the su

The two - digit numbers which when divided by 4, yield 1 as remainder, are as follows: -

13, 17, … 97

Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P. with first term 13 and common difference 4.

Last Term of the A.P(l) = 97

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 -

Sn = (n/2)[a + l]

S22 = (22/2)[13 + 97]

= 11 × 110

= 1210

Thus, the required sum is 1210.

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