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 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.