# the sum of the first 8 terms of a geometric sequence is 6560. The common ratio is 3. What is the third term of the sequence?

the sum of the first 8 terms of a geometric sequence is 6560. The common ratio is 3. What is the third term of the sequence?
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hesgidiauE

The formula for the sum of nn terms of a geometric sequence is ${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{\left(1-r\right)}$ where ${a}_{1}$ is the first term and rr is the common ratio.
If the sum of the first $n=8$ terms is ${S}_{8}=6560$ and the common ratio is $r=3$, then:
${S}_{8}=\frac{{a}_{1}\left(1-{3}^{8}\right)}{1-3}$
$6560=\frac{{a}_{1}\left(1-6561\right)}{-2}$
$6560=\frac{\frac{-6560{a}_{1}}{-2}}{}$
$6560=-3280{a}_{1}$
$-2={a}_{1}$
Now that we know ${a}_{1}$, we can find the third term. The nnth term of a geometric sequence is given by the formula ${a}_{n}={a}_{1}{r}^{n}-1$.
Since ${a}_{1}=-2,r=3,$ and $n=3$ for the third term, then:
${a}_{3}=-2{\left(3\right)}^{3}-1=-2{\left(3\right)}^{2}=-2\left(9\right)=-18$