Question 1092277
<br>Using a for the first term and d for the common difference...<br>
S(10) = a + (a+d) + (a+2d) + ... + (a+9d) = 10a+45d
S(50) = a + (a+d) + (a+2d) + ... + (a+49d) = 50a+1225d<br>
So we know
{{{10a+45d = 50}}}
{{{50a+1225d = 0}}}<br>
Multiply the first equation by 5 and subtract from the second equation to eliminate variable a, and solve for d.<br>
{{{50a+225d = 250}}}
{{{1000d = -250}}}
{{{d = -250/1000 = -1/4}}}<br>
The common difference, d, is -1/4.  Plug this value into an earlier equation to solve for a.<br>
{{{10a+45(-1/4) = 50}}}
{{{10a-45/4 = 200/4}}}
{{{10a = 245/4}}}
{{{a = 245/40 = 49/8}}}<br>
The first term, a, is 49/8.  The common difference, d, is -1/4.<br>
We are asked to find the 10th term, and the sum of the first 100 terms.<br>
The 10th term is the first term, plus the common difference 9 times:
{{{49/8 + 9(-1/4) = 49/8-18/8 = 31/8}}}<br>
The 10th term is 31/8.<br>
The sum of the first 100 terms is 100 times the average of the first and last (100th) terms.  The 100th term is the first term, plus the common difference 99 times:<br>
{{{49/8 + 99(-1/4) = 49/8 - 198/8 = -149/8}}}<br>
The average of the first and 100th terms is<br>
{{{(49/8 + -149/8)/2 = -100/16 = -25/4}}}<br>
And finally the sum of the first 100 terms is<br>
{{{100(-25/4) = -625}}}