College Algebra, Chapter 9, 9.3, Section 9.3, Problem 42

Suppose the second and fifth terms of a geometric sequence are $10$ and $1250$, respectively. Is $31,250$ a term of this sequence? If so, which term is it?

Since this sequence is geometric, its formula is given by $a_n = ar^{n-1}$. Thus,

$a_2 = ar^{2-1} = ar$

$a_5 = ar^{5-1} = ar^4$

From the values we are given for these two terms, we get the following system of equations


$
\left\{
\begin{equation}
\begin{aligned}

10 =& ar
\\
1250 =& ar^4

\end{aligned}
\end{equation}
\right.
$


We solve this system by dividing


$
\begin{equation}
\begin{aligned}

\frac{ar^4}{ar} =& \frac{1250}{10}
&&
\\
\\
r^3 =& 125
&& \text{Simplify}
\\
\\
r =& 5
&& \text{Take the cube root of each side}

\end{aligned}
\end{equation}
$


Substituting $r =5$ into the first equation $ar = 10$, gives


$
\begin{equation}
\begin{aligned}

5a =& 10
\\
a =& 2
\qquad \text{Divide by } 5

\end{aligned}
\end{equation}
$


It follows that the $n$th term of this sequence is,

$a_n = 2(5)^{n-1}$

If $a_n = 31,250$, then


$
\begin{equation}
\begin{aligned}

31,250 =& 2(5)^{n-1}
&&
\\
\\
\frac{31,250}{2} =& (5)^{n-1}
&& \text{Divide by } 2
\\
\\
15625 =& (5)^{n-1}
&& \text{Simplify}
\\
\\
\ln 15625 =& \ln 5^{n-1}
&& \text{Take $\ln$ of each side}
\\
\\
\ln 15625 =& (n-1) \ln 5
&& \text{Property of } \ln
\\
\\
\frac{\ln 15625}{\ln 5} =& n-1
&& \text{Divide by } \ln 5
\\
\\
n =& \frac{\ln 15625}{\ln 5} + 1
&& \text{Add } 1
\\
\\
n =& 7
&&

\end{aligned}
\end{equation}
$


So $31,250$ is on the 7th term.