What is the sum of a 22-term arithmetic sequence where the first term is 56 and the last term is -28?

1 Answer
A. S. Adikesavan
Apr 10, 2016

308

Explanation:

The sequence is {56+(n-1)d}, 2 = 1, 2, 3, ...,22, wher d is the common difference.

The last term (n=22) is -28=56+21d−28=56+21d,
So, d=-82/21=-4d=−8221=−4.

The sum = (22)(56)+(d+2d+3d,...+21d)=1232-(21)(22/2)(-4). using 1+2+3...+n=(n(n+1))/2
The sum simplifies to 308.

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