Answer
$0.7488$
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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Given $a_{2}=0.12,\ a_{5}=0.00096$, we find r:
$\displaystyle \frac{a_{5}}{a_{2}}=\frac{ar^{4}}{ar^{1}}=r^{3}$.
$r^{3}=\displaystyle \frac{a_{5}}{a_{2}}=\frac{0.00096}{0.12}=0.008=(0.2)^{3}$
$r=0.2$.
Now that we know r, we find a from $a_{2}=a\cdot r^{1}$.
$a=\displaystyle \frac{a_{2}}{r}=\frac{0.12}{0.2}=0.6$
We can now use the formula for $S_{4}$ ($n=4$).
$S_{4}=(0.6)\displaystyle \times\frac{1-(0.2)^{4}}{1-0.2}=0.7488$.
