## 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 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 $1.1$, representing 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:

$$\require{AMSsymbols} \require{color} \color{red}u_{n} = u_{1}r^{n-1}$$

where $\require{color} \color{red}u_{1}$ is the first term and $\color{red}r$ the common ratio, given by

$$\require{AMSsymbols} \require{color} \color{red}\dfrac{u_{n+1}}{u_{n}} = r$$

## 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.

$$\require{AMSsymbols} \require{color} \color{red} \dfrac{u_{2}}{u_{1}} = \dfrac{u_{3}}{u_{2}} = \dfrac{u_{4}}{u_{3}} = \cdots = r$$

## 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 $1\ 000\ 000$.

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