Question

find sum of all integral multiples of 3 between 4 and 97.
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Answer 1

Answer:

The n-th term of an arithmetic series = a(n) = a1 + (n-1)d, where d is the common difference

So the formula for the series is a(n) = 9 + (n-1)*3 = 6 + 3n

The last term = 828

For the last term n = (828 - 6)/3 = 274

The sum of an arithmetic series = S(n) = (n/2)(a1+a(n)) = (274/2)(9+828) = 114669