Question

Find the sum of two digit nymber which when ddivided by 3 yields 1 as remainder

Answer 1

first 2 digit number when divided by 3 leaves remainder 1 = 10

last 2 digit number when divided by 3 leaves remainder 1 = 97

a = 10; d = 3

number of terms

97 = 10 + (n-1)3

97 - 10 = (n-1)3

87/3 = n-1

29=n-1

n = 30

Sn =

$= \frac{n}{2} (2a + (n - 1)d) \\ = \frac{30}{2} (2 \times 10) + (29 \times 3) \\ = 15(20 + 87) \\ = 15 \times 107 \\ = 1605$