Mastering Arithmetic and Geometric Series for Mathematics Examinations

What the examiner is testing

Examiners assess your ability to model real-world scenarios using arithmetic and geometric progressions, specifically requiring you to derive unknown terms or sums from given information. Marks are awarded for correctly identifying the type of series, formulating the correct equations from the problem statement, and accurately solving for the required values.

The method

1. Identify the type of series: Determine if the sequence has a common difference (arithmetic) or a common ratio (geometric). Look for phrases like "adds the same amount each time" or "multiplies by the same factor".
2. Extract known values: List the first term \(a\), the common difference \(d\) or common ratio \(r\), the number of terms \(n\), the \(n\)-th term \(u_n\), or the sum of \(n\) terms \(S_n\).
3. Select the appropriate formula:
   • For arithmetic series: \(u_n = a + (n-1)d\) or \(S_n = \frac{n}{2}(2a + (n-1)d)\) or \(S_n = \frac{n}{2}(a + l)\) where \(l\) is the last term.
   • For geometric series: \(u_n = ar^{n-1}\) or \(S_n = \frac{a(1-r^n)}{1-r}\) (for \(|r| < 1\)) or \(S_n = \frac{a(r^n-1)}{r-1}\) (for \(|r| > 1\)).
   • For sum to infinity of a geometric series: \(S_\infty = \frac{a}{1-r}\) (only if \(|r| < 1\)).
4. Formulate and solve equations: Substitute the known values into the chosen formula(s) and solve for the unknown quantity. This may involve solving simultaneous equations or logarithmic equations.
5. Contextualise and check: Ensure your answer makes sense in the context of the problem. For example, \(n\) must be a positive integer.

Worked example

A sequence is defined by \(u_n = 5n - 2\). Find the 10th term and the sum of the first 10 terms.

The sequence is arithmetic as it has the form \(an+b\).
The first term \(a\) is \(u_1 = 5(1) - 2 = 3\).
The common difference \(d\) is \(u_2 - u_1 = (5(2) - 2) - 3 = 8 - 3 = 5\).

To find the 10th term, \(u_{10}\):
$$u_{10} = a + (10-1)d$$
$$u_{10} = 3 + (9) \times 5$$
$$u_{10} = 3 + 45$$
$$u_{10} = 48$$

To find the sum of the first 10 terms, \(S_{10}\):
$$S_{10} = \frac{n}{2}(2a + (n-1)d)$$
$$S_{10} = \frac{10}{2}(2(3) + (10-1)5)$$
$$S_{10} = 5(6 + 9 \times 5)$$
$$S_{10} = 5(6 + 45)$$
$$S_{10} = 5(51)$$
$$S_{10} = 255$$

Sanity check: The terms are 3, 8, 13, ..., 48. The average term is \((3+48)/2 = 25.5\). There are 10 terms, so the sum is \(10 \times 25.5 = 255\). This matches.

Worked example: a harder one

A geometric series has first term \(a\) and common ratio \(r\). The sum of the first four terms is 15. The sum to infinity is 16. Find the possible values of \(a\) and \(r\).

The obvious first move is to write down the formulas:
$$S_4 = \frac{a(1-r^4)}{1-r} = 15 \quad (*)$$
$$S_\infty = \frac{a}{1-r} = 16 \quad (**)$$

Substituting \((**)\) into \((*)\):
$$16(1-r^4) = 15$$
$$1-r^4 = \frac{15}{16}$$
$$r^4 = 1 - \frac{15}{16}$$
$$r^4 = \frac{1}{16}$$
$$r = \pm \sqrt[4]{\frac{1}{16}}$$
$$r = \pm \frac{1}{2}$$

Now substitute these values of \(r\) back into \((**)\) to find \(a\).

Case 1: \(r = \frac{1}{2}\)
$$\frac{a}{1 - \frac{1}{2}} = 16$$
$$\frac{a}{\frac{1}{2}} = 16$$
$$2a = 16$$
$$a = 8$$

Case 2: \(r = -\frac{1}{2}\)
$$\frac{a}{1 - (-\frac{1}{2})} = 16$$
$$\frac{a}{1 + \frac{1}{2}} = 16$$
$$\frac{a}{\frac{3}{2}} = 16$$
$$\frac{2a}{3} = 16$$
$$2a = 48$$
$$a = 24$$

The possible pairs of \((a, r)\) are \((8, \frac{1}{2})\) and \((24, -\frac{1}{2})\).
The initial move of writing down the formulas for \(S_4\) and \(S_\infty\) was correct, but the "twist" is that \(r\) can be negative, leading to two sets of solutions, which students often miss. Also, the condition \(|r|<1\) for \(S_\infty\) is satisfied by both \(r = \frac{1}{2}\) and \(r = -\frac{1}{2}\).

Practice

1. An arithmetic series has first term 7 and common difference 3. Find the 15th term.
2. The first term of a geometric series is 100, and the common ratio is 0.8. Find the sum of the first 8 terms, giving your answer to 3 significant figures.
3. A geometric series has third term 36 and sixth term 4.5. Find the first term and the common ratio.
4. The first three terms of an arithmetic series are \(2p\), \(5p-1\), and \(3p+5\).
   (a) Find the value of \(p\).
   (b) Find the sum of the first 20 terms of the series.

Answers

1. \(u_{15} = 7 + (15-1)3 = 7 + 14 \times 3 = 7 + 42 = 49\)
2. \(S_8 = \frac{100(1-0.8^8)}{1-0.8} = \frac{100(1-0.16777216)}{0.2} = \frac{100(0.83222784)}{0.2} = 500 \times 0.83222784 = 416.11392 \approx 416\) (3 s.f.)
3. \(u_3 = ar^2 = 36\)
   \(u_6 = ar^5 = 4.5\)
   Divide the sixth term by the third term:
   $$\frac{ar^5}{ar^2} = \frac{4.5}{36}$$
   $$r^3 = \frac{1}{8}$$
   $$r = \sqrt[3]{\frac{1}{8}}$$
   $$r = \frac{1}{2}$$
   Substitute \(r = \frac{1}{2}\) into \(ar^2 = 36\):
   $$a\left(\frac{1}{2}\right)^2 = 36$$
   $$a\left(\frac{1}{4}\right) = 36$$
   $$a = 36 \times 4$$
   $$a = 144$$
   So, \(a = 144\) and \(r = \frac{1}{2}\).
4. (a) For an arithmetic series, the common difference is constant.
   $$(5p-1) - 2p = (3p+5) - (5p-1)$$
   $$3p-1 = 3p+5-5p+1$$
   $$3p-1 = -2p+6$$
   $$5p = 7$$
   $$p = \frac{7}{5}$$
   (b) First term \(a = 2p = 2\left(\frac{7}{5}\right) = \frac{14}{5}\).
   Common difference \(d = 3p-1 = 3\left(\frac{7}{5}\right)-1 = \frac{21}{5}-1 = \frac{21-5}{5} = \frac{16}{5}\).
   Sum of the first 20 terms \(S_{20}\):
   $$S_{20} = \frac{20}{2}\left(2\left(\frac{14}{5}\right) + (20-1)\left(\frac{16}{5}\right)\right)$$
   $$S_{20} = 10\left(\frac{28}{5} + 19\left(\frac{16}{5}\right)\right)$$
   $$S_{20} = 10\left(\frac{28}{5} + \frac{304}{5}\right)$$
   $$S_{20} = 10\left(\frac{332}{5}\right)$$
   $$S_{20} = 2 \times 332$$
   $$S_{20} = 664$$

The three mistakes that lose marks

1. Confusing arithmetic and geometric series formulas: Using \(ar^{n-1}\) for an arithmetic sequence or \(a+(n-1)d\) for a geometric sequence. This leads to completely incorrect values. For instance, finding the 5th term of the arithmetic sequence 2, 4, 6... using \(ar^{n-1}\) with \(a=2, r=2\) gives \(2 \times 2^4 = 32\), instead of the correct arithmetic value \(2 + (5-1)2 = 10\).
2. Incorrectly applying the sum to infinity condition: Calculating \(S_\infty\) for a geometric series where \(|r| \ge 1\). This yields a finite value when the series actually diverges. For example, if \(a=1, r=2\), calculating \(S_\infty = \frac{1}{1-2} = -1\), which is mathematically meaningless in this context as the sum grows infinitely large.
3. Errors with negative common ratios: When \(r\) is negative, students often forget to include the negative sign when calculating powers, especially for \(r^n\). For example, if \(r = -2\) and \(n=3\), \(r^3 = (-2)^3 = -8\), not \(2^3 = 8\). This sign error propagates through calculations, leading to an incorrect sum or term.

30-second recap

Arithmetic series have a common difference \(d\); use \(u_n = a+(n-1)d\) and \(S_n = \frac{n}{2}(2a+(n-1)d)\). Geometric series have a common ratio \(r\); use \(u_n = ar^{n-1}\) and \(S_n = \frac{a(1-r^n)}{1-r}\). Remember \(S_\infty = \frac{a}{1-r}\) only when \(|r|<1\).