Question 1150465
<br>
t(1) = 11/3; t(3) = 3 = 9/3.<br>
Let d be the common difference.<br>
{{{t(3) = t(1)+2d}}}
{{{9/3 = 11/3+2d}}}
{{{-2/3 = 2d}}}
{{{d = -1/3}}}<br>
{{{S(n) = n((t(1)+t(n))/2)}}}<br>
{{{t(n) = t(1)+(n-1)d}}}
{{{t(n) = 11/3+(n-1)(-1/3) = 11/3-n/3+1/3 = (12-n)/3}}}<br>
{{{S(n) = n((11/3+(12-n)/3))/2 = n((23-n)/3)/2 = n(23-n)/6}}}
{{{20 = n(23-n)/6}}}
{{{120 = n(23-n)}}}
{{{n^2-23n+120 = 0}}}
{{{(n-8)(n-15) = 0}}}<br>
There are two solutions to the problem: n = 8 and n = 15.<br>
The sum of terms 1 through 8 is 20; then the sum of terms 9 through 15 is 0, so the sum of terms 1 through 15 is again 20.<br>