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 an = 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, Sn = n/2 [2a + (n - 1)d]
Therefore,
S22 = 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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