Geometric Sequence

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

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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