SOLUTION: Let a_1, a_2, a_3, ... be an arithmetic sequence. Let S_n denote the sum of the first n terms. If S_{10} = 1 and S_{200} = 1/2, then find S_{15}.

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Question 1209385: Let a_1, a_2, a_3, ... be an arithmetic sequence. Let S_n denote the sum of the first n terms. If S_{10} = 1 and S_{200} = 1/2, then find S_{15}.
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Write the terms of the sequence in terms of the first term a and common difference d:

a, a+d, a+2d, a+3d, ...

The sum of n terms of the sequence is

sum = (number of terms) * (average of all terms)

And in an arithmetic sequence the average of all the terms is the average of the first and last terms:

sum = (number of terms) * (average of first and last terms)

For the sum of the first 10 terms, the 10th term is a+9d; the sum is



For the sum of the first 200 terms, the 200th term is a+199d; the sum is



The sum of the first 10 terms is 1, and the sum of the first 200 terms is 1/2:

-->

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This is a straightforward problem of solving a pair of linear equations in two variables, but the numbers are ugly. Use pencil and paper if you want; or solve the pair of equations using wolframalpha.com to find

first term: 7951/76000
common difference: -39/38000

Then use those to find

ANSWER: the sum of the first 15 terms is

The corrected solution was verified using an excel spreadsheet:

first term: 7951/76000
common difference: -39/38000
sum of first 10 terms: 1
sum of first 15 terms: 4443/3040
sum of first 200 terms: 1/2
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
My ugly answer differs from Greenestamps' ugly answer, although we
agree with the ugly first terms and common difference.

Let a_1, a_2, a_3, ... be an arithmetic sequence. Let S_n denote the sum of the
first n terms. If S_{10} = 1 and S_{200} = 1/2, then find S_{15}.








Multiply the first equation by -40 so the a1 terms will cancel:









Go back to



Multiply the first equation by -7960 and the second equation by 9 so
the terms in d will cancel.







Substituting in











Now we'll substitute n=15











Edwin
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