find the sum to terms of an AS whose 8th term is 42 and 16th term is 82
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The AS can be written a, a+d, a+2d, ..., a+(n-1)d where a is the first term and d the common difference.

The sum to n terms is na+d(1+2+3+...+(n-1))=an+dn(n-1)/2.

When n=8, a+7d=42 and when n=16, a+15d=82, so 8d=40 and d=5; a=42-35=82-75=7.

The sum to n terms is 7n+5n(n-1)/2=(14n+5n^2-5n)/2=n(5n+9)/2.

The series is 7, 12, 17, 22, 27, 32, 37, 42, ..., 82, ...

Put n=3 sum to 3 terms=3*24/2=36=7+12+17, which checks out OK.

by Top Rated User (1.2m points)

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