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}.

Algebra ->  Sequences-and-series -> 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}.      Log On


   

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) About Me  (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

10%28%282a%2B9d%29%2F2%29

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

200%28%282a%2B199d%29%2F2%29

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

10%28%282a%2B9d%29%2F2%29=1 --> 2a%2B9d=1%2F5

200%28%282a%2B199d%29%2F2%29=1%2F2 --> 2a=199d=1%2F200

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 cross%281481%2F7600%29 4443%2F3040

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) About Me  (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}.


system%28%0D%0A%0D%0A1=5%282a%5B1%5D%2B9d%29%2C%0D%0A1%2F2=100%282a%5B1%5D%2B199d%29%29

system%28%0D%0A%0D%0A1=10a%5B1%5D%2B45d%2C%0D%0A1%2F2=200a%5B1%5D%2B19900d%29

system%28%0D%0A%0D%0A1=10a%5B1%5D%2B45d%2C%0D%0A1=400a%5B1%5D%2B39800d%29

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

system%28%0D%0A%0D%0A-40=-400a%5B1%5D-1800d%2C%0D%0A1=400a%5B1%5D%2B39800d%29

-39=38000d

%28-39%29%2F38000=d

d=-39%2F38000

Go back to

system%28%0D%0A%0D%0A1=10a%5B1%5D%2B45d%2C%0D%0A1=400a%5B1%5D%2B39800d%29

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

system%28%0D%0A%0D%0A-7960=-79600a%5B1%5D-358200d%2C%0D%0A9=3600a%5B1%5D%2B358200d%29

-7951=-76000a%5B1%5D
%28-7951%29%2F%28-76000%29=a%5B1%5D
a%5B1%5D=7951%2F76000

Substituting in

S%5Bn%5D=expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

S%5Bn%5D=expr%28n%2F2%29%282%287951%2F76000%29%2B%28n-1%29%28-39%2F3800%29%29

S%5Bn%5D=expr%28n%2F2%29%287951%2F38000-%2839n%29%2F3800%2B39%2F3800%29%29

S%5Bn%5D=expr%28n%2F2%29%287951%2F38000-%28390n%29%2F38000%2B390%2F38000%29%29

S%5Bn%5D=expr%28n%2F2%29%28%288341-390n%29%2F38000%29

Now we'll substitute n=15

S%5B15%5D=expr%2815%2F2%29%28%288341-390%2815%29%29%2F38000%29

S%5B15%5D=expr%2815%2F2%29%28%288341-5850%29%2F38000%29

S%5B15%5D=expr%2815%2F2%29%282491%2F38000%29

S%5B15%5D=expr%283%2F2%29%282491%2F7600%29

S%5B15%5D=7473%2F15200

Edwin