SOLUTION: Find a<sub>n</sub> for the arithmetic series with S<sub>16</sub> = -288, and a<sub>1</sub> = -21

Algebra ->  Sequences-and-series -> SOLUTION: Find a<sub>n</sub> for the arithmetic series with S<sub>16</sub> = -288, and a<sub>1</sub> = -21      Log On


   

Question 1092307: Find an for the arithmetic series
with S16 = -288, and a1 = -21
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

The sum of n terms of an arithmetic series is the number of terms, n, multiplied by the average of all the terms. Since the numbers in an arithmetic series are equally spaced, the average of all the terms is the average of the first and last.

If S_16 = -288, then the average of all the terms is -288 divided by 16, which is -18.

If a_1 is -21 and the average of the 16 terms is -18, then the last term a_16 is -15.

a_16 is a_1 plus the common difference 15 times:
-15+=+-21+%2B+15d
6+=+15d
d+=+6%2F15+=+2%2F5

The first term is -21 and the common difference is 2/5; the formula for the n-th term is
a%28n%29+=+-21+%2B+%282%2F5%29%28n-1%29

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
S%5Bn%5D%22%22=%22%22expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%5E%22%22%29

Substitute n = 16, a1 = -21

S%5B16%5D%22%22=%22%22expr%2816%2F2%29%282%28-21%29%2B%2816-1%29d%5E%22%22%29

Substitute S16 = -288.

-288%22%22=%22%22%288%29%282%28-21%29%2B%2816-1%29d%5E%22%22%29 

Now simplify that and solve for d.

When you find d, substitute it and a1 = -21

into 

a%5Bn%5D%22%22=%22%22a%5B1%5D%2B%28n-1%29d

And simplify the right side:

You finish.

Edwin