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# In a series -9,-6,-3,----, find the number of terms n to get a sum $S_{n}$ = 66.

$\begin{array}{1 1} -4 \\ 11\\ 8 \\ 6\end{array}$

Explanation: Observe the series -9,,-6,-3-------
It is an AP with a=-9 ,d=+3
In an AP,sum $S_{n}\;$ is given by
$S_{n}=\frac{n}{2}[2a+(n-1)d]=66$
$\frac{n}{2}[2*(-9)+3(n-1)]=66$
$\frac{n}{2}[(-18)+3n-3)]=66$
$3n^2-21n-132=0$
$3n^2+12n-33n-132=0$
$3n(n+4)-33(n+4)=0$
$3(n-11)(n+4)=0$
$n-11=0$
$n=11$
$or\;n+4=0$
$n=-4$
Since number of terms in a series cannot be negative, answer n=11 is valid.