The General Term for a Geometric Sequence

From the flu example above we know that \({T}_{1}=2\) and \(r=2\), and we have seen from the table that the \(n\)\(^{\text{th}}\) term is given by \({T}_{n}=2\times {2}^{n-1}\).

The general geometric sequence can be expressed as:

\[\begin{array}{rll} T_{1} &=a &=ar^{0} \\ T_{2} &=a \times r &=ar^{1} \\ T_{3} &=a \times r \times r &=ar^{2} \\ T_{4} &=a \times r \times r \times r &=ar^{3} \\ T_{n} &=a \times [r \times r \ldots (n-1)\text{ times}] &=ar^{n-1} \end{array}\]

Therefore the general formula for a geometric sequence is:

\[T_{n} = ar^{n-1}\]

where

\(a\) is the first term in the sequence;
\(r\) is the constant ratio.

Test for a geometric sequence

To test whether a sequence is a geometric sequence or not, check if the ratio between any two consecutive terms is constant:

\[\cfrac{T_{2}}{T_{1}} = \cfrac{T_{3}}{T_{2}} = \cfrac{T_{n}}{T_{n-1}} = r\]

If this condition does not hold, then the sequence is not a geometric sequence.

Example

Question

We continue with the previous flu example, where \(T_{n}\) is the number of newly-infected people after \(n\) days:

\[T_{n} = 2 \times 2^{n-1}\]

Calculate how many newly-infected people there are on the tenth day.
On which day will \(\text{16 384}\) people be newly-infected?

Write down the known values and the general formula

\begin{align*} a &= 2 \\ r &= 2 \\ T_{n} &= 2 \times 2^{n-1} \end{align*}

Use the general formula to calculate \(T_{10}\)

Substitute \(n = 10\) into the general formula:

\begin{align*} {T}_{n}& = {a} \times {r}^{n-1} \\ \therefore {T}_{10}& = 2\times {2}^{10-1} \\ & = 2\times {2}^{9} \\ & = 2\times 512 \\ & = 1024 \end{align*}

On the tenth day, there are \(\text{1 024}\) newly-infected people.

Use the general formula to calculate \(n\)

We know that \(T_{n} = \text{16 384}\) and can use the general formula to calculate the corresponding value of \(n\):

\begin{align*} {T}_{n}& = {a} {r}^{n-1} \\ \text{16 384}& = 2 \times {2}^{n-1} \\ \cfrac{\text{16 384}}{2} & = {2}^{n-1} \\ \text{8 192}& = {2}^{n-1} \\ \text{We can write } \text{8 192} &\text{ as } {2}^{13} \\ \text{So } {2}^{13}& = {2}^{n-1} \\ \therefore 13& = n-1 \quad (\text{same bases})\\ \therefore n & = 14 \end{align*}

There are \(\text{16 384}\) newly-infected people on the \(14^{\text{th}}\) day.

For this geometric sequence, plotting the number of newly-infected people (\(T_{n}\)) vs. the number of days (\(n\)) results in the following graph:

Day (n)	No. of newly-infected people
\(\text{1}\)	\(2\)
\(\text{2}\)	\(\text{4}\)
\(\text{3}\)	\(\text{8}\)
\(\text{4}\)	\(\text{16}\)
\(\text{5}\)	\(\text{32}\)
\(\text{6}\)	\(\text{64}\)
\(n\)	\(2 \times 2^{n-1}\)

In this example we are only dealing with positive integers \(( n \in \{1; 2; 3; \ldots \}, T_{n} \in \{1; 2; 3; \ldots \} )\), therefore the graph is not continuous and we do not join the points with a curve (the dotted line has been drawn to indicate the shape of an exponential graph).

Geometric mean

The geometric mean between two numbers is the value that forms a geometric sequence together with the two numbers.

For example, the geometric mean between \(\text{5}\) and \(\text{20}\) is the number that has to be inserted between \(\text{5}\) and \(\text{20}\) to form the geometric sequence: \(5; x; 20\)

\begin{align*} \text{Determine the constant ratio: } \cfrac{x}{5} &= \cfrac{20}{x} \\ \therefore x^{2} &= 20 \times 5 \\ x^{2} &= 100 \\ x &= \pm 10 \end{align*}

Important: remember to include both the positive and negative square root. The geometric mean generates two possible geometric sequences:

\[5; 10; 20; \ldots\]\[5; -10; 20; \ldots\]

In general, the geometric mean (\(x\)) between two numbers \(a\) and \(b\) forms a geometric sequence with \(a\) and \(b\):

\[\text{For a geometric sequence: }a; x; b\]\begin{align*} \text{Determine the constant ratio: } \cfrac{x}{a} &= \cfrac{b}{x} \\ x^{2} &= ab \\ \therefore x &= \pm \sqrt{ab} \end{align*}

The General Term for a Geometric Sequence