Question 1136301
<br>
Let a = first term
let d = common difference<br>
Then 4th term = a+3d
and 9th term = a+8d<br>
Sum of 4th and 9th terms = 58 =(a+3d)+(a+8d)<br>
{{{2a+11d = 58}}}  (1)<br>
Sum of first 26 terms = 390 = 26*average of 1st and 26th terms<br>
{{{26((a+a+25d)/2)=390}}}
{{{2a+25d = 30}}}  (2)<br>
From (1) and (2)...<br>
{{{2a+25d = 30}}}
{{{2a+11d = 58}}}
{{{14d = -28}}}
{{{d = -2}}}<br>
Then<br>
{{{2a+11(-2) = 58}}}
{{{2a-22 = 58}}}
{{{2a = 80}}}
{{{a = 40}}}<br>
The first term is 40; the common difference is -2.<br>
With first term 40 and common difference -2, the 21st term will be 40-40 = 0 and the 41st term will be 40-80 = -40.  That means the sum of the first 81 terms will be 0, because all the positive terms at the beginning of the sequence get cancelled out by later negative terms.<br>
So the smallest n for which the sum of n terms of the sequence is negative is 82.