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We are asked to find the eighth term of a geometric sequence. Given that $a_{3}=81$ and $r=3,$ we can find the first term of the sequence using the equation for the $n_{th}$ term of the sequence.
The first term of the sequence is $9,$ then the equation for the $n_{th}$ term of the sequence becomes as follows.
$a_{n}=a_{1}⋅3_{n−1}⇒a_{n}=9⋅3_{n−1} $
Now, we are ready to find the eighth term, $a_{8}.$
The eight term of the sequence is $19683.$

$a_{n}=a_{1}⋅r_{n−1}$

SubstituteIIISubstitute $n=3,a_{3}=81,r=3$

$81=a_{1}⋅3_{3−1}$

Solve for $a_{1}$

SubTermSubtract term

$81=a_{1}⋅3_{2}$

CalcPowCalculate power

$81=a_{1}⋅9$

DivEqn$LHS/9=RHS/9$

$9=a_{1}$

RearrangeEqnRearrange equation

$a_{1}=9$