A sequence \(\{u_n\}\) is a geometric progression (GP) if \[ \frac{u_{n+1}}{u_n} = r \quad \text{for all } n, \] where \(r\) is a constant called the common ratio.

Notation:

• First term: \(u_1 = a\).
• Common ratio: \(r\).
• General term: \[ u_n = ar^{\,n-1}, \quad n = 1,2,3,\dots \]

Example

GP: \(3,\, 6,\, 12,\, 24,\, \dots\)

• \(a = 3\)
• \(r = \dfrac{6}{3} = 2\)
• \(u_n = 3 \cdot 2^{n-1}\), e.g. \(u_4 = 3 \cdot 2^{3} = 24\).

Let \(S_n\) be the sum of the first \(n\) terms of a GP: \[ S_n = u_1 + u_2 + \dots + u_n = a + ar + ar^2 + \dots + ar^{n-1}. \]

For \(r \neq 1\), using the standard derivation (multiply by \(r\), subtract and simplify), we obtain:

Example

Find the sum of the first 5 terms of the GP \(5,\, 10,\, 20,\, \dots\).

Here \(a = 5\), \(r = 2\), \(n = 5\). \[ \begin{aligned} S_5 &= 5 \cdot \frac{2^{5} - 1}{2 - 1} = 5(32 - 1) = 5 \times 31 = 155. \end{aligned} \]

So the sum of the first 5 terms is \(\boxed{155}\).

Consider the infinite geometric series \[ a + ar + ar^{2} + ar^{3} + \dots \]

• If \(|r| < 1\), the terms get smaller and the series has a finite sum.
• If \(|r| \ge 1\), the series does not converge (no finite sum to infinity).

When \(|r| < 1\), we take the limit of \(S_n\) as \(n \to \infty\): \[ \begin{aligned} S_n &= a\,\frac{1 - r^{n}}{1 - r}, \\ \text{and } r^{n} &\to 0 \text{ as } n \to \infty. \end{aligned} \]

Example

Find the sum to infinity of the GP \(8,\, 4,\, 2,\, 1,\, \dots\).

Here \(a = 8\), \(r = \dfrac{4}{8} = \dfrac{1}{2}\) and \(|r| = \dfrac{1}{2} < 1\). \[ \begin{aligned} S_{\infty} &= \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16. \end{aligned} \]

So the sum to infinity is \(\boxed{16}\).

If you are given two terms of a GP, you can usually find \(a\) and \(r\) by dividing one term by another.

Example

In a GP, the third term is 18 and the sixth term is \( \dfrac{9}{2} \). Find the values of \(a\) and \(r\), and determine whether the series has a sum to infinity.

Step 1: Use the nth term formula.

\[ u_n = ar^{n-1}. \] Third term: \[ u_3 = ar^{2} = 18. \] Sixth term: \[ u_6 = ar^{5} = \frac{9}{2}. \]

Step 2: Divide to find \(r^3\).

\[ \begin{aligned} \frac{u_6}{u_3} &= \frac{ar^{5}}{ar^{2}} = r^{3} = \frac{\frac{9}{2}}{18} = \frac{9}{2} \cdot \frac{1}{18} = \frac{1}{4}. \end{aligned} \] So \[ r^{3} = \frac{1}{4} \quad \Rightarrow \quad r = \sqrt[3]{\frac{1}{4}} = \frac{1}{\sqrt[3]{4}}. \]

Step 3: Find \(a\).

From \(ar^{2} = 18\), \[ a = \frac{18}{r^{2}}. \] (You can leave \(a\) in exact form using \(r^{2} = \sqrt[3]{\tfrac{1}{16}}\).)

Step 4: Sum to infinity?

Since \(|r| = \left|\sqrt[3]{\frac{1}{4}}\right| < 1\), the series has a sum to infinity given by \[ S_{\infty} = \frac{a}{1 - r}. \] (This can be found if required in an exam question.)

Worked example

A geometric progression has first term \(a\) and common ratio \(r\), where \(|r| < 1\).

1. Show that the sum to infinity of the series is \(\dfrac{a}{1 - r}\).
2. In a particular GP, the sum to infinity is 20 and the sum of the first 3 terms is 18. Find \(a\) and \(r\).

Solution.

(i) Already shown in Section 3:

\[ \begin{aligned} S_n &= a\,\frac{1 - r^{n}}{1 - r} \\ \text{and if } |r| < 1,\, r^{n} &\to 0 \Rightarrow S_{\infty} = \frac{a}{1 - r}. \end{aligned} \]

(ii) Use the given information.

Sum to infinity: \[ S_{\infty} = \frac{a}{1 - r} = 20 \quad\Rightarrow\quad a = 20(1 - r). \]

Sum of first 3 terms: \[ S_3 = a + ar + ar^{2} = a(1 + r + r^{2}) = 18. \] Substitute \(a = 20(1 - r)\): \[ \begin{aligned} 20(1 - r)(1 + r + r^{2}) &= 18, \\ 20(1 - r^{3}) &= 18, \quad\text{(since }(1 - r)(1 + r + r^{2}) = 1 - r^{3}\text{)}\\ 1 - r^{3} &= \frac{18}{20} = \frac{9}{10}, \\ r^{3} &= 1 - \frac{9}{10} = \frac{1}{10}. \end{aligned} \] So \[ r = \sqrt[3]{\frac{1}{10}}. \]

Now \[ a = 20(1 - r) = 20\left(1 - \sqrt[3]{\frac{1}{10}}\right). \]

So \[ \boxed{r = \sqrt[3]{\frac{1}{10}}, \quad a = 20\left(1 - \sqrt[3]{\frac{1}{10}}\right)}. \] (In an exam you may leave the answer in exact form unless a decimal is requested.)