A geometric sequence multiplies by the same number at every step, so it models repeated percentage growth and decay. This note covers the formula for the \(n\)th term, the formula for the sum of the first \(n\) terms, and sigma notation for geometric series. It is the basis for compound interest, population growth and radioactive decay.
Key ideas
A sequence is geometric if each term is found by multiplying the previous term by a fixed number \(r\), called the common ratio:
\[r = \frac{u_{n+1}}{u_n}\]
For example, \(3, 6, 12, 24, \dots\) has \(u_1 = 3\) and \(r = 2\). The sequence \(80, 40, 20, 10, \dots\) has \(r = \tfrac{1}{2}\). A negative ratio makes the terms alternate in sign: \(12, -6, 3, -1.5, \dots\) has \(r = -\tfrac{1}{2}\).
- If \(r \gt 1\) the terms grow in size without limit.
- If \(0 \lt r \lt 1\) the terms shrink towards zero.
- If \(r \lt 0\) the terms alternate between positive and negative.
To get from \(u_1\) to \(u_n\) you multiply by \(r\) a total of \(n-1\) times. This gives \(u_n = u_1 r^{n-1}\).
A geometric series is the sum of terms of a geometric sequence. To find \(S_n\), write the sum, multiply it by \(r\), and subtract. Almost all of the terms cancel, leaving \(S_n(1-r) = u_1(1 - r^n)\). Dividing by \(1-r\) gives the result. The formula is valid for any \(r \ne 1\). The two forms \(\frac{u_1(r^n-1)}{r-1}\) and \(\frac{u_1(1-r^n)}{1-r}\) are the same. Choose the one that avoids negative numbers: use the first when \(r \gt 1\) and the second when \(r \lt 1\).
Finding \(r\) from two terms. If you know \(u_p\) and \(u_q\) with \(q \gt p\), then \(\dfrac{u_q}{u_p} = r^{q-p}\). Take the root to find \(r\). If \(q-p\) is even, there may be both a positive and a negative solution, so check the information in the question.
Sigma notation. The expression \(\sum_{k=1}^{10} 3 \times 2^{k}\) means add the terms \(3 \times 2^1, 3 \times 2^2, \dots, 3 \times 2^{10}\). The first term is \(6\) (not 3), and the ratio is \(2\). Always work out the first term by substituting the first value of \(k\).
When does a term or sum pass a target? Use the table feature on your GDC, or solve the inequality with logarithms once you know how. Check that your \(n\) is a whole number.
A useful link with percentages: increasing by \(p\%\) each step means \(r = 1 + \frac{p}{100}\), and decreasing by \(p\%\) means \(r = 1 - \frac{p}{100}\). So a 20% fall gives \(r = 0.8\) and a 5% rise gives \(r = 1.05\). This is why geometric sequences appear in so many real situations.
Be careful to tell sequences and series apart. "Find the 10th term" asks for \(u_{10}\). "Find the total of the first 10 terms" asks for \(S_{10}\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(r = \dfrac{u_{n+1}}{u_n}\) | Common ratio. Constant for a geometric sequence. |
| \(u_n = u_1 r^{n-1}\) | The \(n\)th term. |
| \(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\) | Sum of the first \(n\) terms, \(r \ne 1\). Convenient when \(r \gt 1\). |
| \(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\) | The same sum. Convenient when \(r \lt 1\). |
| \(\dfrac{u_q}{u_p} = r^{q-p}\) | Finding \(r\) when two terms are known. |
| \(\sum_{k=1}^{n} u_k\) | Sigma notation for the sum of the first \(n\) terms. |
Worked examples
Example 1 A geometric sequence has second term 12 and fifth term 324, and all of its terms are positive. Find the first term, the common ratio and the sum of the first 8 terms.
\(u_2 = u_1 r = 12\) and \(u_5 = u_1 r^4 = 324\).
Divide the second equation by the first: \(\dfrac{u_1 r^4}{u_1 r} = r^3 = \dfrac{324}{12} = 27\), so \(r = 3\).
Then \(u_1 \times 3 = 12\), so \(u_1 = 4\).
\(S_8 = \dfrac{4(3^8 - 1)}{3 - 1} = \dfrac{4 \times 6560}{2} = 13\,120\).
Answer: \(u_1 = 4\), \(r = 3\), \(S_8 = 13\,120\).
Example 2 A factory makes 500 units in week 1. Because a machine is wearing out, its output falls by 20% each week, so the weekly outputs are 500, 400, 320, ... Find the output in week 10 and the total output over the first 10 weeks. Give answers to 3 significant figures.
\(u_1 = 500\) and \(r = 0.8\).
\(u_{10} = 500 \times 0.8^{9} = 500 \times 0.134217\ldots = 67.1\).
Since \(r \lt 1\), use \(S_{10} = \dfrac{500(1 - 0.8^{10})}{1 - 0.8}\).
\(0.8^{10} = 0.107374\ldots\), so \(S_{10} = \dfrac{500 \times 0.892626\ldots}{0.2} = 2231.56\ldots\)
Answer: week 10 output \(\approx 67.1\) units and total output \(\approx 2230\) units.
Example 3 A video is shared by 3 people in the first hour after posting. The number of new shares doubles each hour, so the new shares in successive hours are 3, 6, 12, ... Find the first hour in which the number of new shares exceeds 10 000.
\(u_n = 3 \times 2^{n-1}\). We need \(3 \times 2^{n-1} \gt 10\,000\).
Use the table feature on your GDC to list values of \(u_n\). \(u_{12} = 3 \times 2^{11} = 6144\), which is not enough. \(u_{13} = 3 \times 2^{12} = 12\,288\), which is greater than 10 000.
Answer: the 13th hour.
Common errors
- Common error: using \(u_n = u_1 r^{n}\). The power is \(n-1\).
- Common error: writing \((u_1 r)^{n-1}\) when only \(r\) should be raised to the power. Remember that \(u_1\) is not raised to a power.
- Common error: reading the first term of \(\sum_{k=1}^{10} 3 \times 2^k\) as 3. It is \(3 \times 2^1 = 6\).
- Common error: miscalculating \(r\) as \(u_{n+1} - u_n\). The common ratio is a quotient, not a difference.
- Common error: rounding \(r\) or intermediate powers too early. Keep full calculator values until the end.
Exam tip: When two terms are given, divide one equation by the other to cancel \(u_1\) at once. Then use the information in the question (for example, "all terms are positive") to choose the correct root.
Check yourself
- Find the 8th term of the sequence \(5, 10, 20, \dots\)
Answer
\(r = 2\). \(u_8 = 5 \times 2^{7} = 5 \times 128 = 640\). - Find the sum of the first six terms of \(81, 27, 9, \dots\) to 3 significant figures.
Answer
\(r = \tfrac{1}{3}\). \(S_6 = \dfrac{81\left(1 - (1/3)^6\right)}{1 - 1/3} = \dfrac{81 \times (728/729)}{2/3} = 121.33\ldots \approx 121\). - Evaluate \(\sum_{k=1}^{10} 3 \times (1.5)^{k-1}\) to 3 significant figures.
Answer
\(u_1 = 3\) and \(r = 1.5\). \(S_{10} = \dfrac{3(1.5^{10} - 1)}{0.5} = 6(57.665\ldots - 1) = 339.99\ldots \approx 340\). - For the sequence \(2, 6, 18, \dots\), find the smallest \(n\) for which \(S_n \gt 5000\).
Answer
\(S_n = \dfrac{2(3^n - 1)}{2} = 3^n - 1\). \(S_7 = 2186\) (too small) and \(S_8 = 6560\), which is greater than 5000. So \(n = 8\).