Question

How many terms of the series $-9,-6,-3,...$ must be taken that the sum may be $66$?

Answer 1

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$66=\frac{n}{2}\left[2\right(-9)+(n-1\left)3\right]$

$132=n(3n-21)$

$3n^2-21n-132=0$

$n^2-7n-44=0$

$(n-11)(n+4)=0$

$n=11,-4$

$\therefore n=11$ terms of series must be taken