Question

Find the sum of all 3 digit no . Which leaves the same remainder 2 when divided by 5

Answer 1

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according to question

a=102 , d=5 , l=997

$l = a + (n - 1)d \\ 997 = 102 + (n - 1)5 \\ 895 = (n - 1)5 \\ \frac{895}{5} = n - 1 \\ 179 = n - 1 \\ n = 180$

$s_{180} = \frac{n}{2} \:(a + l) \\ = \frac{180}{2} (102 + 997) \\ = 90(1099) \\ s_{180} = 98910$

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