Question 1092302
<br>You don't say so, but I will assume this is an arithmetic sequence....<br>
In any arithmetic sequence, the sum of n terms is n times the average of the first and last terms.<br>
The sum of the first 20 terms is 20 times the average of the first and 20th terms.  If the first term is a and the common difference is d, then the 20th term is a+19d.  Since your problem says the sum of the first 20 terms is 200,
{{{20(a+(a+19d))/2 = 200}}}
{{{10(2a+19d) = 200}}}
{{{2a+19d = 20}}}  (1)<br>
Also in your problem the sum of the next 10 terms (terms 21 through 30) is 400.  The 21st term is a+20d; the 30th is a+29d.  The sum of those 10 terms is
{{{10((a+20d)+(a+29d))/2 = 400}}}
{{{2a+49d = 80}}}  (2)<br>
Subtract equation (1) from equation (2) and solve for d; then plug that value for d in either (1) or (2) to solve for a.<br>
{{{30d = 60}}}
{{{d = 2}}}
{{{2a+19(2) = 20}}}
{{{2a+38 = 20}}}
{{{2a = -18}}}
{{{a = -9}}}<br>
We are to find the 40th term, which is a+39d:
{{{-9+40(2) = -9+80 = 71}}}<br>
The 40th term is 71.<br>
Oops!  I said the 40th term is a+39d, but then I calculated a+40d.<br>
The 40th term is a+39d:
{{{-9+39(2) = -9+78 = 69}}}<br>
So the 40th term is 69, which is one of your answer choices.