Question 1209385
<pre>
My ugly answer differs from Greenestamps' ugly answer, although we
agree with the ugly first terms and common difference.</pre>
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}.
<pre>
{{{system(S[n]=expr(n/2)(2a[1]+(n-1)d),

1=expr(10/2)(2a[1]+(10-1)d),
1/2=expr(200/2)(2a[1]+(200-1)d))}}}

{{{system(

1=5(2a[1]+9d),
1/2=100(2a[1]+199d))}}}

{{{system(

1=10a[1]+45d,
1/2=200a[1]+19900d)}}}

{{{system(

1=10a[1]+45d,
1=400a[1]+39800d)}}}

Multiply the first equation by -40 so the a<sub>1</sub> terms will cancel:

{{{system(

-40=-400a[1]-1800d,
1=400a[1]+39800d)}}}

{{{-39=38000d}}}

{{{(-39)/38000=d}}}

{{{d=-39/38000}}}

Go back to

{{{system(

1=10a[1]+45d,
1=400a[1]+39800d)}}}

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

{{{system(

-7960=-79600a[1]-358200d,
9=3600a[1]+358200d)}}}

{{{-7951=-76000a[1]}}}
{{{(-7951)/(-76000)=a[1]}}}
{{{a[1]=7951/76000}}}

Substituting in

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

{{{S[n]=expr(n/2)(2(7951/76000)+(n-1)(-39/3800))}}}

{{{S[n]=expr(n/2)(7951/38000-(39n)/3800+39/3800))}}}

{{{S[n]=expr(n/2)(7951/38000-(390n)/38000+390/38000))}}}

{{{S[n]=expr(n/2)((8341-390n)/38000)}}}

Now we'll substitute n=15

{{{S[15]=expr(15/2)((8341-390(15))/38000)}}}

{{{S[15]=expr(15/2)((8341-5850)/38000)}}}

{{{S[15]=expr(15/2)(2491/38000)}}}

{{{S[15]=expr(3/2)(2491/7600)}}}

{{{S[15]=7473/15200}}}

Edwin</pre>