College Algebra, Chapter 9, 9.3, Section 9.3, Problem 46

Determine the partial sum $S_n$ of the geometric sequence that satisfies $\displaystyle a_2 = 0.12, a_5 = 0.00096$ and $n=4$.

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

0.12 =& ar
\\
0.00096 =& ar^4

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


We solve this system by dividing.


$
\begin{equation}
\begin{aligned}

\frac{ar^4}{ar} =& \frac{0.00096}{0.12}
&&
\\
\\
r^3 =& 0.008
&& \text{Simplify}
\\
\\
r =& 0.2
&& \text{Take the cube root of each side}

\end{aligned}
\end{equation}
$


Substituting for $r$ in the first equation $ar = 0.12$, gives


$
\begin{equation}
\begin{aligned}

0.12 =& a(0.2)
\\
\\
a =& \frac{0.12}{0.2}
\qquad \text{Divide by } 0.2
\\
\\
a =& \frac{3}{5}
\\
\\
a =& 0.6

\end{aligned}
\end{equation}
$


Then using the formula for partial sum


$
\begin{equation}
\begin{aligned}

S_n =& a \frac{1 - r^n}{1-r}
\\
\\
S_4 =& 0.6 \left( \frac{1-0.2^4}{1-0.2} \right)
\\
\\
S_4 =& 0.7488

\end{aligned}
\end{equation}
$