 # Geometric Series Formulas

Geometric Series are series where you find the next term by multiplying the previous term with a constant quotient $k$. They can be used to make a model of many different situations, from the development of a bacterial culture to loans and savings. What’s nice about geometric series is that we can learn everything about one of them through the following three formulas:

Formula

### GeometricSeries

Finding the quotient in the series:

 $k=\frac{{a}_{n+1}}{{a}_{n}}$

Finding the $n$th term of a series:

 ${a}_{n}={a}_{1}\cdot {k}^{n-1}$

Finding the sum of the series:

If $k=1$, you use this formula to find the sum:

 ${S}_{n}={a}_{1}\cdot n$

Example 1

You have the geometric series

 $3+9+27+81+\cdots \phantom{\rule{0.17em}{0ex}}.$

Find the quotient, an expression for the $n$th term, and the sum of the first 10 terms.

In exercises like this, you can just fill out the formulas. The quotient $k$ is

 $k=\frac{{a}_{n+1}}{{a}_{n}}=\frac{9}{3}=3.$

The expression for the $n$th term is

 ${a}_{n}={a}_{1}\cdot {k}^{n-1}=3\cdot {3}^{n-1}={3}^{n}.$

That makes the sum of the first 10 terms

 ${S}_{10}=3\cdot \frac{{3}^{10}-1}{3-1}=88\phantom{\rule{0.17em}{0ex}}572.$

Example 2

Find ${a}_{1}$ and $k$ when you know that ${a}_{3}=4$ and ${a}_{5}=16$ are two terms in an increasing geometric series.

As long as you are given two different terms, it can be smart to solve the exercise just like you would solve equations with two unknowns. You know that the formula for an arbitrary term in a geometric series is given by ${a}_{n}={a}_{1}\cdot {k}^{n-1}$.

That gives you

$\begin{array}{llll}\hfill {a}_{3}& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}\cdot {k}^{3-1}& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}\cdot {k}^{2}& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}& =\frac{4}{{k}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill {a}_{5}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}\cdot {k}^{5-1}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}\cdot {k}^{4}& =16,\phantom{\rule{1em}{0ex}}{a}_{1}=\frac{4}{{k}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ⇒\phantom{\rule{-0.17em}{0ex}}\left(\frac{4}{{k}^{2}}\right)\cdot {k}^{4}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 4{k}^{2}& =16\phantom{\rule{1em}{0ex}}|:4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {k}^{2}& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =±\sqrt{4}=±2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill ⇒{a}_{1}& =\frac{4}{{\left(±2\right)}^{2}}=1.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llllllll}\hfill {a}_{3}& =4\phantom{\rule{2em}{0ex}}& \hfill {a}_{5}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}\cdot {k}^{3-1}& =4\phantom{\rule{2em}{0ex}}& \hfill {a}_{1}\cdot {k}^{5-1}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}\cdot {k}^{2}& =4\phantom{\rule{2em}{0ex}}& \hfill {a}_{1}\cdot {k}^{4}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}& =\frac{4}{{k}^{2}}\phantom{\rule{2em}{0ex}}& \hfill \phantom{\rule{-0.17em}{0ex}}\left(\frac{4}{{k}^{2}}\right)\cdot {k}^{4}& =16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill 4{k}^{2}& =16\phantom{\rule{1em}{0ex}}|:4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill {k}^{2}& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill k& =±\sqrt{4}=±2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}_{1}& =\frac{4}{{\left(±2\right)}^{2}}=1.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Because we know the series is increasing, $k=-2$ can’t be the correct quotient, making it a false solution. That gives us ${a}_{1}=1$ and $k=2$.