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

**Solution:**

The two-digit numbers which when divided by 4, yields 1 as remainder are 13,17, 21, ...., 97 .

This 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 a^{n} = a + (n - 1) d

Therefore,

97 = 13 + (n - 1)(4)

4(n - 1) = 97 - 13

n - 1 = 84/4

⇒ n = 22

Sum of n terms of an A.P, S_{n} = n/2 [2a + (n - 1)d]

Therefore,

S_{22} = 222/2 [2(13) + (22 - 1)(4)]

= 11 [26 + 84]

= 11 x 110

= 1210

Thus, the required sum is 1210

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 6

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

**Summary:**

Therefore the sum of all two-digit numbers when divided by 4, gives 1 as a remainder is 1210

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