A geometric sequence is also referred as a geometric progression. Each term of a geometric sequence can be obtained from the previous one my 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, this is 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.

### Question 2

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

### Question 3

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

### Question 4

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