Answer
$-645$
Work Step by Step
See p. 861.
For the geometric sequence $a_{n}=ar^{n-1}$
the nth partial sum$ S_{n}=\displaystyle \sum_{k=1}^{n}ar^{k-1}$ (where $r\neq 1$)
is given by$ \displaystyle \quad S_{n}=a\cdot\frac{1-r^{n}}{1-r}$
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We see that $a=-15,\quad r=-2$
So, $a_{n}=ar^{n-1} =-15(-2)^{n-1}$
Given the last term, we find n:
$-960=-15(-2)^{n-1}\qquad/\div(-15)$
$64=(-2)^{n-1}$
... recognizing $64$ as a power of 2,
.... or factoring, $64$=2($32$)=$2^{2}(16)...=2^{6}=(-2)^{6}$
$(-2)^{6}=(-2)^{n-1}$
$n-1=6$
$n=7$
So
$S_{7}=-15\displaystyle \cdot\frac{1-(-2)^{7}}{1-(-2)}=-15\cdot\frac{1+128}{3}$
$=-5(129)=-645$
