# Geometric Sequence and Geometric Mean

## Geometric Sequence Definition

Geometric Sequences are sequences where each term is obtained by multiplying the preceding term by a certain constant factor, which is often called $\textit{common ratio}$. A geometric sequence is also referred to as a $\textit{geometric progression}$.

David expects $10$% increase per month to deposit to his account. A $10$% increase per month would mean that the amount would increase by a constant factor of $(1+0.1)$ or $1.1$.
He starts off with $100$ initially.
By the beginning of the second month he will expect $$100 \times 1.1 = 110$$, by the start of the third month he would expect $$110 \times 1.1 = 100 \times 1.1^2 = 121$$, and so on.
This is an example of a geometric sequence

The first term is $10$, and the common factor is $1.1$, which represents a $10\%$ increase on the previous term.

\begin{align} \displaystyle u_{n+1} &= u_{n} \times 1.1 \\ u_{1} &= \100 \\ u_{2} &= \100 \times 1.1 \\ u_{3} &= \100 \times 1.1^2 \\ u_{4} &= \100 \times 1.1^3 \\ &\cdots \\ u_{n} &= \100 \times 1.1^{n-1} \\ \end{align}

For a geometric sequence:
$$\require{color} \color{red}u_{n} = u_{1}r^{n-1}$$
where $\require{color} \color{red}u_{1}$ is the first term and $\color{red}r$ the common ratio, given by
$$\require{color} \color{red}\dfrac{u_{n+1}}{u_{n}} = r$$

## Geometric Mean

If we consider three consecutive terms in a geometric sequence $\{x,y,z\}$ then
$$\dfrac{y}{x} = \dfrac{z}{y} = r$$
where $r$ is the common ratio.
Thus the middle term, $y$, called the $\textit{geometric mean}$, can be calculated in terms of the outer two terms, $x$ and $z$.
$$y^2 = xz$$
A geometric sequence is a sequence of numbers for which the ratio of successive terms is the same.
$$\require{color} \color{red} \dfrac{u_{2}}{u_{1}} = \dfrac{u_{3}}{u_{2}} = \dfrac{u_{4}}{u_{3}} = \cdots = r$$

## Example 1

Determine whether the sequence $\{2,10, 50, 250,\cdots\}$ is geometric.

\begin{align} \displaystyle \dfrac{u_{2}}{u_{1}} &= \dfrac{10}{2} = 5 \\ \dfrac{u_{3}}{u_{2}} &= \dfrac{50}{10} = 5 \\ \dfrac{u_{4}}{u_{3}} &= \dfrac{250}{50} = 5 \\ \end{align}
There is a common ratio of $5$.
Therefore this is a geometric sequence.

## Example 2

Determine whether the sequence $\{4,-8,16,-32,64,\cdots\}$ is geometric.

\begin{align} \displaystyle \dfrac{u_{2}}{u_{1}} &= \dfrac{-8}{4} = -2 \\ \dfrac{u_{3}}{u_{2}} &= \dfrac{16}{-8} = -2 \\ \dfrac{u_{4}}{u_{3}} &= \dfrac{-32}{16} = -2 \\ \end{align}
There is a common ratio of $-2$.
Therefore this is a geometric sequence.

## Example 3

Find the $n$th term and the $12$th in the geometric sequence where the first term is $3$ and the fourth term is $-24$.

\begin{align} \displaystyle u_{1} &= 3 \\ u_{4} &= -24 \\ u_{1} \times r^{4-1} &= -24 \\ 3 \times r^{3} &= -24 \\ r^{3} &= -8 \\ r^{3} &= (-2)^3 \\ \therefore r &= -2 \\ u_{n} &= 3 \times (-2)^{n-1} \\ u_{12} &= 3 \times (-2)^{12-1} \\ &= 3 \times (-2)^{11} \\ &= -6144 \end{align}

## Example 4

The first three terms of a geometric sequence are $2, 6$ and $18$. Which numbered term would be the first to exceed $1\ 000\ 000$ in this sequence?

\begin{align} \displaystyle u_{1} &= 2 \\ r &= \dfrac{6}{2} = 3 \\ u_{n} &= 2 \times 3^{n-1} \\ 2 \times 3^{n-1} &\gt 1\ 000\ 000 \\ 3^{n-1} &\gt 500\ 000 \\ n-1 &\gt \log_{3}{500\ 000} \\ n &\gt \log_{3}{500\ 000} + 1 \\ &\gt \dfrac{\log_{10}{500\ 000}}{\log_{10}{3}} + 1 \\ &\gt 12.94 \cdots \\ \end{align}
The $13$th term would be the first to exceed $1\ 000\ 000$. 