Answer

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**Hint:**Assuming the terms are nonzero, we can find the common ratio $r$on a calculator by taking any two consecutive terms and dividing the later one by the earlier one.

**Complete step by step solution:**

A geometric sequence is a sequence with a common ratio $r$ between adjacent terms,

That is a sequence of the form ${a_1}$, ${a_1}$ $r$, ${a_1}$$r_{}^2$, …${a_1}$$r_{}^n$

Then assuming the terms are nonzero, dividing any term by the prior term give the ratio:

$\Rightarrow$ $\dfrac{{{a_1}r_{}^n}}{{{a_1}r_{}^{n - 1}}}$

Cancelling ${a_1}$from numerator and denominator we get,

$\Rightarrow$ $\dfrac{{r_{}^n}}{{r_{}^{n - 1}}}$

Now by law of exponential $\dfrac{{a_{}^m}}{{a_{}^n}}$= $a_{}^{m - n}$

We can write $\dfrac{{r_{}^n}}{{r_{}^{n - 1}}}$ as

$\Rightarrow$ $r_{}^{n - (n - 1)}$

$\Rightarrow$ $r_{}^{n - n + 1}$

$\Rightarrow$ $r_{}^1$ = $r$

To find $r$ on a calculator, then take any two consecutive terms and divide the later one by the earlier one.

So, $r$= $\dfrac{{{a_{n + 1}}}}{{{a_n}}}$.

**Note:**

More generally, given any two terms ${a_1}r_{}^m$ and${a_1}r_{}^n$, $m < n$

We can find $r$by dividing $\dfrac{{{a_1}r_{}^n}}{{{a_1}r_{}^m}}$ and taking the ${(n - m)^{th}}$ root:

$\Rightarrow$ $(\dfrac{{{a_1}r_{}^n}}{{{a_1}r_{}^m}})_{}^{\dfrac{1}{{n - m}}}$ = $(r_{}^{n - m})_{}^{\dfrac{1}{{n - m}}}$ = $r_{}^{\dfrac{{n - m}}{{n - m}}}$ = $r_{}^1$= $r$.

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