SOLUTION: In an arithmetic series t1 = 11/3 , t3 = 3. Find the value of n for which Sn = 20.

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Question 1150465: In an arithmetic series t1 = 11/3 , t3 = 3. Find the value of n for which Sn = 20.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


t(1) = 11/3; t(3) = 3 = 9/3.

Let d be the common difference.

t%283%29+=+t%281%29%2B2d
9%2F3+=+11%2F3%2B2d
-2%2F3+=+2d
d+=+-1%2F3

S%28n%29+=+n%28%28t%281%29%2Bt%28n%29%29%2F2%29

t%28n%29+=+t%281%29%2B%28n-1%29d
t%28n%29+=+11%2F3%2B%28n-1%29%28-1%2F3%29+=+11%2F3-n%2F3%2B1%2F3+=+%2812-n%29%2F3


20+=+n%2823-n%29%2F6
120+=+n%2823-n%29
n%5E2-23n%2B120+=+0
%28n-8%29%28n-15%29+=+0

There are two solutions to the problem: n = 8 and n = 15.

The sum of terms 1 through 8 is 20; then the sum of terms 9 through 15 is 0, so the sum of terms 1 through 15 is again 20.