Answer
please see step-by-step
Work Step by Step
A geometric sequence has the form
$a, ar, ar^{2}, ar^{3}, \ldots$
The number $a$ is the first term of the sequence, and the number $r $is the common ratio.
The nth term of the sequence is $\quad a_{n}=ar^{n-1}$
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For $a_{1},\ a_{2},\ a_{3},\ \ldots$
let the common ratio be $r_{1}$, so
$a_{n}=a_{1}r_{1}^{n-1}$ .
For $b_{1},\ b_{2},\ b_{3},\ \ldots$
let the common ratio be $r_{2}$., so
$b_{n}=b_{1}r_{2}^{n-1}$.
Observing the sequence $a_{1}b_{1},\ a_{2}b_{2},\ a_{3}b_{3},\ \ldots$
$a_{n}b_{n}=a_{1}r_{1}^{n-1}\cdot b_{1}r_{2}^{n-1}=(a_{1}b_{1})\cdot(r_{1}r_{2})^{n-1}$.
It is geometric, with first term $a_{1}b_{1}$
and common ratio $r_{1}r_{2}$.
