Question 1092219
<br>We have an arithmetic sequence with
S(10) = 50
S(50) = 0<br>
Since the sum of the first 10 terms is 50 and the sum of the first 50 terms is 0, we know that the sum of terms 11 through 50 must be -50:<br>
A(11)+A(12)+...+A(50) = -50<br>
Now each of the terms 11 through 20 is 10 common differences greater than the corresponding terms 1 through 10, so if the common difference is d,
A(11)+A(12)+...+A(20) = 50+100d<br>
Similarly,
A(21)+A(22)+...+A(30) = 50+200d
A(31)+A(32)+...+A(40) = 50+300d
A(41)+A(42)+...+A(50) = 50+400d<br>
Adding all these together, we have
A(11)+A(12)+...+a(50) = 200+1000d<br>
But we know this sum is -50, so
{{{200+1000d = -50}}}
{{{1000d = -250}}}
{{{d = -1/4}}}<br>
So the common difference in the sequence is -1/4.<br>
The sum of the first 10 terms is
A(1)+(A(1)+d)+(A(1)+2d)+...+(A(1)+9d) = 10A(1)+45d<br>
The sum of the first 10 terms is 50, so
{{{10A(1)+45d = 10A(1)+45(-1/4) = 50}}}
{{{10A(1)-45/4 = 200/4}}}
{{{10A(1) = 245/4}}}
{{{A(1) = 245/40 = 49/8}}}<br>
So now we have the first term (49/8) and the common difference (-1/4).<br>
We are asked to find the value of A(10), which is A(1)+9d:
{{{A(10) = 49/8 + 9(-1/4) = 49/8 - 18/8 = 31/8}}}<br>
The 10th term is 31/8.<br>
We are also asked to find S(100).  S(100) is 100 times the average of A(1) and A(100).  A(100) is the first term plus 99 times the common difference.<br>
{{{A(100) = 49/8 + 99(-1/4) = 49/8 - 198/8 = -149/8}}}
{{{S(100) = 100((49/8 - 149/8)/2) = 100(-100/16) = -625}}}<br>
The sum of the first 100 terms is -625.