Geometric Sequences - Extended Maths

The common ratio

A geometric sequence grows or shrinks by multiplying by a fixed number each time. That number is the common ratio, \(r\). To find it, divide any term by the one before it.

Common ratio: The fixed multiplier between consecutive terms: \(r = \dfrac{\text{term}}{\text{previous term}}\). If it is the same for every pair, the sequence is geometric.

For \(3,\ 6,\ 12,\ 24,\ \ldots\) each term is double the last, so \(r = \tfrac{6}{3} = 2\). For \(8,\ 4,\ 2,\ 1,\ \ldots\) each term is half the last, so \(r = \tfrac{4}{8} = \tfrac{1}{2}\). A ratio can be negative too: \(2,\ -6,\ 18,\ \ldots\) has \(r = -3\).

Check the ratio is really constant

Test more than one pair. \(\tfrac{6}{3}=2\) and \(\tfrac{12}{6}=2\) and \(\tfrac{24}{12}=2\): all equal, so it is geometric. If the ratios differ, the sequence is not geometric and these formulas do not apply.

The nth term

Call the first term \(a\). To reach the 2nd term you multiply by \(r\) once, the 3rd term twice, and so on. The \(n\)th term needs \(n-1\) multiplications.

\[ n\text{th term} = a\,r^{\,n-1} \]

Worked example

Find the common ratio and the 6th term of \(3,\ 6,\ 12,\ 24,\ \ldots\)

Divide a term by the one before it: \(r = \tfrac{6}{3} = 2\). The first term is \(a = 3\).

Write the rule: \(n\text{th term} = 3 \cdot 2^{\,n-1}\).

Put \(n = 6\): \(3 \cdot 2^{5} = 3 \cdot 32 = 96\).

\(r = 2\) and the 6th term \(= 96\).

Sum of n terms

Adding up the first \(n\) terms would be slow by hand, so we use a formula. Either version below works; pick whichever keeps the numbers positive.

\[ S_n = \frac{a\,(1 - r^{\,n})}{1 - r} = \frac{a\,(r^{\,n} - 1)}{r - 1} \qquad (r \neq 1) \]

Worked example

Find the sum of the first 8 terms of \(3,\ 6,\ 12,\ 24,\ \ldots\)

Here \(a = 3\), \(r = 2\), \(n = 8\). Since \(r > 1\), use \(S_n = \dfrac{a(r^{n}-1)}{r-1}\).

Work out \(r^{n} = 2^{8} = 256\).

Substitute: \(S_8 = \dfrac{3\,(256 - 1)}{2 - 1} = \dfrac{3 \cdot 255}{1}\).

\(S_8 = 765\).

Sum to infinity

If the terms keep shrinking, that is when \(|r| < 1\), the running total closes in on a fixed value rather than growing forever. That value is the sum to infinity.

\[ S_\infty = \frac{a}{1 - r} \qquad \text{valid only when } |r| < 1 \]

Worked example

Find the sum to infinity of \(8 + 4 + 2 + 1 + \ldots\)

First term \(a = 8\); ratio \(r = \tfrac{4}{8} = \tfrac{1}{2}\). Since \(|r| = \tfrac{1}{2} < 1\), the sum to infinity exists.

Substitute into \(S_\infty = \dfrac{a}{1-r}\): \(S_\infty = \dfrac{8}{1 - \tfrac{1}{2}} = \dfrac{8}{\tfrac{1}{2}}\).

Dividing by \(\tfrac{1}{2}\) is the same as multiplying by 2: \(8 \times 2 = 16\).

\(S_\infty = 16\).

Always check the condition first

The sum to infinity only exists when \(|r| < 1\). If \(|r| \geq 1\) the terms do not shrink and there is no finite total, so write "no sum to infinity". Stating and justifying this condition is exactly the reasoning credited under Criterion C (communicating) and Criterion D (applying maths).

Check yourself

1. Find the common ratio and the 5th term of \(5,\ 15,\ 45,\ \ldots\) +

Ratio \(r = \tfrac{15}{5} = 3\) and first term \(a = 5\). The rule is \(5 \cdot 3^{\,n-1}\). For the 5th term put \(n = 5\): \(5 \cdot 3^{4} = 5 \cdot 81 = \mathbf{405}\).

2. Find the sum of the first 6 terms of \(2,\ 6,\ 18,\ \ldots\) +

Here \(a = 2\), \(r = 3\), \(n = 6\). Using \(S_n = \dfrac{a(r^{n}-1)}{r-1}\) with \(3^{6} = 729\): \(S_6 = \dfrac{2\,(729 - 1)}{3 - 1} = \dfrac{2 \cdot 728}{2} = \mathbf{728}\).

3. Find the sum to infinity of \(27 + 9 + 3 + \ldots\) +

First term \(a = 27\); ratio \(r = \tfrac{9}{27} = \tfrac{1}{3}\). Since \(|r| = \tfrac{1}{3} < 1\), the sum exists: \(S_\infty = \dfrac{27}{1 - \tfrac{1}{3}} = \dfrac{27}{\tfrac{2}{3}} = 27 \times \tfrac{3}{2} = \mathbf{40.5}\).

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