Answer
441
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 $a_{3}=28,\ a_{6}=224$, we find r:
$\displaystyle \frac{a_{6}}{a_{3}}=\frac{ar^{5}}{ar^{2}}=r^{3}$.
$r^{3}=\displaystyle \frac{a_{6}}{a_{3}}=\frac{224}{28}=8$,
$r=2$.
Now that we know r, we find a from $a_{3}=a\cdot r^{2}$.
$a=\displaystyle \frac{a_{3}}{r^{2}}=\frac{28}{2^{2}}=7$.
We can now use the formula for $S_{6}$ ($n=6$).
$S_{6}=7\displaystyle \frac{1-2^{6}}{1-2}$
$=(-7)(-63)$
$=441$.
