# Everything You Need to Know about Geometric Sequences

A geometric sequence is also referred to as a geometric progression. Each term of a geometric sequence can be obtained from the previous one by multiplying by the same non-zero constant.

For example, $2, \ 6, \ 18, \ 54, \cdots$ is a geometric sequence as each term can be obtained by multiplying the previous term by $3$. Notice that $6 \div 2 = 18 \div 6 = 54 \div 18 = 3$, so each term divided by the previous one gives the same constant, often called a common ratio.

## Algebraic Definition of a Geometric Sequence

$\dfrac{T_{n+1}}{T_{n}} = r$ for all positive integers $n$ where $r$ is a constant called the common ratio.

## An Important Property of a Geometric Sequence

If $a, \ b$ and $c$ are any consecutive terms of a geometric sequence then $\dfrac{b}{a} = \dfrac{c}{b}$.
That is, $b^2 = ac$ and so $b = \pm \sqrt{ac}$ where $\sqrt{ac}$ is the geometric mean of $a$ and $c$.

## The General Term Formula of a Geometric Sequence

Suppose the first term of a geometric sequence is $a$, and the common ratio is $r$.
\begin{aligned} \displaystyle T_1 &= a \\ T_2 &= T_1 \times r = ar \\ T_3 &= T_2 \times r = ar \times r = ar^2 \\ T_4 &= T_3 \times r = ar^2 \times r = ar^3 \\ &\text{and so on …} \\ T_n &= T_{n-1} \times r = ar^{n-2} \times r = ar^{n-1} \\ \therefore T_n &= ar^{n-1} \end{aligned}

## Practice Questions

### Question 1

Show that the sequence $240,120,60,30, \cdots$ is geometric.

\begin{aligned} \displaystyle &\dfrac{120}{240} = \dfrac{1}{2} \\ &\dfrac{120}{240} = \dfrac{1}{2} \\ &\dfrac{120}{240} = \dfrac{1}{2} \\ &\dfrac{120}{240} = \dfrac{1}{2} \\ &\text{Consecutive terms have a common ratio of } \dfrac{1}{2}. \\ &\text{Therefore, the sequence is geometric.} \end{aligned}

### Question 2

Find the general term of a geometric sequence: $3, 6, 12, 24, \cdots$.

\begin{aligned} \displaystyle &\dfrac{6}{3} = \dfrac{12}{6} = \dfrac{24}{12} = 2 \\ &\text{the common ratio } r = 2 \\ &\therefore T_n = 3 \times 2^{n-1} \end{aligned}

### Question 3

Find $k$, if $k+1, 3k,$ and $5k+2$ consecutive terms of a geometric sequence.

\begin{aligned} \displaystyle \dfrac{3k}{k+1} &= \dfrac{5k+2}{3k} \\ (3k)^2 &= (k+1) \times (5k+2) \\ 9k^2 &= 5k^2 + 7k + 2 \\ 4k^2-7k-2 &= 0 \\ (k-2)(4k+1) &= 0 \\ \therefore k &= 2 \text{ or } k= -\dfrac{1}{4} \end{aligned}

### Question 4

Find the general term of a geometric sequence; its third term is 48, and its sixth term is -3072.

\begin{aligned} \displaystyle T_3 &= ar^2 = 48 \cdots (1)\\ T_6 &= ar^5 = -3072 \cdots (2)\\ (2) &\div (1) \\ \dfrac{ar^5}{ar^2} &= \dfrac{-3072}{48} \\ r^3 &= -64 \\ r^3 &= (-4)^3 \\ r &= -4 \\ \text{Substitute } r &= 4 \text{ into } T_3 = ar^2 = 48 \cdots (1)\\ a \times 4^2 &= 48 \\ a &= 3 \\ \therefore T_n &= 3 \times (-4)^{n-1} \end{aligned}

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