2.3 Geometric Sequences and Series
SL 1.3
- what a geometric sequence is, and the common ratio \(r\)
- the formula for the \(n\)th term, \(u_n = u_1 r^{n-1}\)
- how to write a geometric sequence as a recurrence relation, and how to show a sequence is geometric
- the formula for the sum of the first \(n\) terms, and why it works
Geometric sequences
A sequence is geometric if each term is the previous term multiplied by the same number. So \(u_1, u_2, u_3, \ldots\) is geometric if there is a number \(r\) with
\[ \frac{u_2}{u_1} = \frac{u_3}{u_2} = \frac{u_4}{u_3} = \cdots = r, \qquad r \neq 0 . \]
The number \(r\) is the common ratio. (A geometric sequence is also called a geometric progression, or GP.)
| Sequence | First term \(u_1\) | Common ratio \(r\) |
|---|---|---|
| \(5, 10, 20, 40, \ldots\) | \(5\) | \(2\) |
| \(81, 27, 9, 3, \ldots\) | \(81\) | \(\frac{1}{3}\) |
| \(3, -6, 12, -24, \ldots\) | \(3\) | \(-2\) |
| \(k, k^3, k^5, k^7, \ldots\) | \(k\) | \(k^2\) |
To find \(r\), always calculate a term divided by the term before it. A negative ratio makes the signs alternate.
The \(n\)th term
Start at \(u_1\) and multiply by \(r\) again and again:
\[ u_1, \quad u_2 = u_1 r, \quad u_3 = u_1 r^2, \quad u_4 = u_1 r^3, \quad \ldots \]
To reach the \(n\)th term we multiply by \(r\) exactly \((n - 1)\) times.
The \(n\)th term of a geometric sequence In the formula booklet
\[ u_n = u_1\,r^{\,n-1} \]
- \(u_n\): the \(n\)th term (general term)
- \(u_1\): the first term
- \(r\): the common ratio
- \(n\): the position of the term (\(n = 1, 2, 3, \ldots\))
Example 1 Find the 20th term of the geometric sequence \(-8,\; 4,\; -2,\; 1,\; \ldots\) Give your answer as a fraction.
\(u_1 = -8\) and \(r = \dfrac{4}{-8} = -\dfrac{1}{2}\).
\[ u_{20} = -8 \times \left(-\frac{1}{2}\right)^{19} = -8 \times \left(-\frac{1}{2^{19}}\right) = \frac{8}{2^{19}} = \frac{1}{2^{16}} = \frac{1}{65\,536} \]
An odd power of a negative number is negative, so the two minus signs cancel.
When two terms are given, divide one equation by the other to remove \(u_1\).
Example 2 The second term of a geometric sequence is \(6\) and the fifth term is \(162\). Find the first term and the common ratio. Hence find the eighth term.
\[ u_2 = u_1 r = 6 \qquad (1) \qquad\qquad u_5 = u_1 r^4 = 162 \qquad (2) \]
\((2) \div (1)\): \(\;r^3 = 27\), so \(r = 3\). Then \(u_1 = \dfrac{6}{3} = 2\).
\[ u_8 = 2 \times 3^7 = 4374 \]
If the question says the terms are positive and you get \(r^2 = \ldots\) (an even power), take only the positive root.
Geometric sequences as recurrence relations
The sequence \(3, 6, 12, 24, \ldots\) doubles each time, so
\[ u_1 = 3, \qquad u_{n+1} = 2u_n . \]
In general, a geometric sequence with first term \(u_1\) and common ratio \(r\) is
\[ u_{n+1} = r\,u_n, \qquad n \in \mathbb{Z}^+ . \]
To show that a sequence is geometric
Find \(\dfrac{u_{n+1}}{u_n}\) and show that it is a constant (it does not depend on \(n\)). That constant is \(r\).
Example 3 A sequence has general term \(u_n = 5 \times 2^{\,n+1}\), \(n \in \mathbb{Z}^+\). Show that the sequence is geometric, and find its first term and common ratio.
\[ \frac{u_{n+1}}{u_n} = \frac{5 \times 2^{\,n+2}}{5 \times 2^{\,n+1}} = 2^{(n+2)-(n+1)} = 2 . \]
This is constant, so the sequence is geometric with \(r = 2\). The first term is \(u_1 = 5 \times 2^2 = 20\).
Geometric series
Sum of the first \(n\) terms of a geometric sequence In the formula booklet
\[ S_n = \frac{u_1\left(r^n - 1\right)}{r - 1} = \frac{u_1\left(1 - r^n\right)}{1 - r}, \qquad r \neq 1 \]
- \(S_n\): the sum of the first \(n\) terms
- \(u_1\): the first term
- \(r\): the common ratio
- \(n\): the number of terms
Both forms give the same answer. The first is tidier when \(r > 1\), the second when \(r < 1\), because then the numerator and denominator are both positive.
Why the formula works
Write \(S_n\), then multiply every term by \(r\):
\[ \begin{aligned} S_n &= u_1 + u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1} \\ r S_n &= \phantom{u_1 + {}} u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1} + u_1 r^{n} \end{aligned} \]
Almost all the terms are the same. Subtract the second line from the first:
\[ S_n - rS_n = u_1 - u_1 r^{n} \;\Longrightarrow\; S_n (1 - r) = u_1 (1 - r^n) \;\Longrightarrow\; S_n = \frac{u_1 (1 - r^n)}{1 - r} . \]
(If \(r = 1\), every term equals \(u_1\) and simply \(S_n = n u_1\).)
Example 4 (a) Find the sum of the first 10 terms of the geometric series with first term \(3\) and common ratio \(2\).
(b) A geometric series has common ratio \(\dfrac{1}{3}\), and the sum of its first 8 terms is \(6560\). Find the first term.
(a) \(S_{10} = \dfrac{3\left(2^{10} - 1\right)}{2 - 1} = 3 \times 1023 = 3069\).
(b) \(1 - \left(\frac{1}{3}\right)^8 = 1 - \dfrac{1}{6561} = \dfrac{6560}{6561}\) and \(1 - \dfrac{1}{3} = \dfrac{2}{3}\), so
\[ S_8 = u_1 \times \frac{6560}{6561} \times \frac{3}{2} = 6560 \;\Longrightarrow\; u_1 = \frac{6561 \times 2}{3} = 4374 . \]
Example 5 The first term and the last term of a geometric series are \(u_1\) and \(l\), and the common ratio is \(r\).
(a) Show that the sum of the series is \(S = \dfrac{u_1 - rl}{1 - r}\).
(b) Hence find the sum \(2 - 6 + 18 - \cdots + 1458\).
(a) If there are \(n\) terms, the last term is \(l = u_1 r^{n-1}\), so \(u_1 r^{n} = r \times u_1 r^{n-1} = rl\). Then
\[ S = \frac{u_1 - u_1 r^{n}}{1 - r} = \frac{u_1 - rl}{1 - r} . \]
(b) \(u_1 = 2\), \(r = -3\), \(l = 1458\):
\[ S = \frac{2 - (-3)(1458)}{1 - (-3)} = \frac{2 + 4374}{4} = 1094 . \]
This avoids finding the number of terms.
Example 6 A sequence is defined by \(u_n = 3\left(\dfrac{2}{5}\right)^{n-1}\), \(n = 1, 2, 3, \ldots\)
(a) Show that the sequence is geometric.
(b) Find the sum of the first \(2n\) terms.
(a) \(\dfrac{u_{n+1}}{u_n} = \dfrac{3\left(\frac{2}{5}\right)^{n}}{3\left(\frac{2}{5}\right)^{n-1}} = \dfrac{2}{5}\), a constant. So it is geometric with \(u_1 = 3\) and \(r = \dfrac{2}{5}\).
(b) Use the formula with \(2n\) terms in place of \(n\):
\[ S_{2n} = \frac{3\left(1 - \left(\frac{2}{5}\right)^{2n}\right)}{1 - \frac{2}{5}} = 5\left(1 - \left(\frac{4}{25}\right)^{n}\right). \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(u_n = u_1 r^n\) | \(u_n = u_1 r^{\,n-1}\) | From \(u_1\) to \(u_n\) you multiply by \(r\) only \((n - 1)\) times. |
| \(u_n = (u_1 r)^{n-1}\) | \(u_1 \times r^{\,n-1}\) | Only \(r\) is raised to the power. |
| \(r = u_1 \div u_2\) | \(r = u_2 \div u_1\) | Later term divided by earlier term. |
| \(\left(-\frac{1}{2}\right)^{19}\) treated as positive | It is negative | An odd power keeps the minus sign. |
| Subtracting terms to find \(r\) | Divide | Subtracting gives \(d\) for an arithmetic sequence. |
Exercises
1 Find the common ratio and the 8th term of \(2, 6, 18, 54, \ldots\)
\(r = 3\). \(\;u_8 = 2 \times 3^7 = 4374\).
2 Find the 10th term of \(96, -48, 24, -12, \ldots\) as a fraction.
\(r = -\dfrac{1}{2}\). \(\;u_{10} = 96 \times \left(-\dfrac{1}{2}\right)^{9} = -\dfrac{96}{512} = -\dfrac{3}{16}\).
3 Which term of the sequence \(5, 10, 20, \ldots\) is \(5120\)?
\(5 \times 2^{n-1} = 5120 \Rightarrow 2^{n-1} = 1024 = 2^{10} \Rightarrow n = 11\). It is the 11th term.
4 The 3rd term of a geometric sequence is \(18\) and the 6th term is \(486\). Find \(u_1\) and \(r\).
\(\dfrac{u_1 r^5}{u_1 r^2} = r^3 = \dfrac{486}{18} = 27\), so \(r = 3\) and \(u_1 = \dfrac{18}{9} = 2\).
5 The 2nd term of a geometric sequence of positive terms is \(12\) and the 4th term is \(3\). Find \(u_1\) and \(r\).
\(r^2 = \dfrac{3}{12} = \dfrac{1}{4}\). The terms are positive, so \(r = \dfrac{1}{2}\) (not \(-\dfrac{1}{2}\)). Then \(u_1 = 24\).
6 The numbers \(x - 2\), \(x + 1\) and \(3x + 3\) are consecutive terms of a geometric sequence. Find the possible values of \(x\).
The ratios are equal: \(\dfrac{x + 1}{x - 2} = \dfrac{3x + 3}{x + 1}\), so \((x + 1)^2 = (x - 2)(3x + 3)\).
\[ x^2 + 2x + 1 = 3x^2 - 3x - 6 \;\Longrightarrow\; 2x^2 - 5x - 7 = 0 \;\Longrightarrow\; (2x - 7)(x + 1) = 0 . \]
\(x = -1\) makes the second term \(0\), which is not allowed in a geometric sequence. So \(x = \dfrac{7}{2}\) (terms \(1.5, 4.5, 13.5\); ratio \(3\)).
7 Show that the sequence \(u_n = 7 \times 3^{\,2n}\) is geometric, and state \(u_1\) and \(r\).
\(\dfrac{u_{n+1}}{u_n} = \dfrac{7 \times 3^{\,2n+2}}{7 \times 3^{\,2n}} = 3^2 = 9\), a constant. \(\;u_1 = 7 \times 9 = 63\), \(\;r = 9\).
8 Find the sum of the first 8 terms of \(1 + 3 + 9 + 27 + \cdots\)
\(S_8 = \dfrac{1\left(3^8 - 1\right)}{3 - 1} = \dfrac{6560}{2} = 3280\).
9 Find the sum \(64 + 32 + 16 + \cdots + \dfrac{1}{4}\).
\(u_1 = 64\), \(r = \dfrac{1}{2}\), \(l = \dfrac{1}{4}\). Using Example 5:
\[ S = \frac{64 - \frac{1}{2} \times \frac{1}{4}}{1 - \frac{1}{2}} = 2\left(64 - \frac{1}{8}\right) = 128 - \frac{1}{4} = 127.75 . \]
10 A geometric series has \(u_1 = 5\) and \(r = 2\). Find the least number of terms needed for the sum to exceed \(10\,000\).
\(S_n = 5\left(2^n - 1\right) > 10\,000 \Rightarrow 2^n - 1 > 2000 \Rightarrow 2^n > 2001\).
\(2^{10} = 1024\) and \(2^{11} = 2048\), so the least number of terms is \(n = 11\).