In this chapter: sequences as functions on N, series and sigma notation, a quick review of AP, the nth term and sum of a GP, sum of an infinite GP, arithmetic and geometric means, the AM ≥ GM inequality, and the sums of n, n² and n³.Sequences and series
A sequence is an ordered list of numbers a1, a2, a3, ..., where an is the nth term. Formally it is a function whose domain is N (an infinite sequence) or {1, 2, ..., k} (a finite sequence). A sequence may be given by a formula, like an = n² + 1, or by a recurrence, like the Fibonacci sequence: a1 = a2 = 1 and an = an−1 + an−2 for n > 2, giving 1, 1, 2, 3, 5, 8, ....
Adding the terms gives a series: a1 + a2 + ... + an, written compactly as ∑ak (k from 1 to n). We write Sn for the sum of the first n terms. A sequence whose terms follow a definite pattern is called a progression.
Arithmetic progression (review)
In an AP each term exceeds the previous one by a fixed common difference d.
The arithmetic mean of a and b is A = (a + b)/2, and a, A, b form an AP. To insert n numbers between a and b so that the whole list is an AP, use d = (b − a)/(n + 1). Adding a constant to every term of an AP, or multiplying every term by a non-zero constant, gives another AP.
Geometric progression
A sequence of non-zero numbers is a GP if the ratio of each term to the previous one is a fixed number r, the common ratio (r ≠ 0). With first term a:
Use the first form of Sn when r > 1 and the second when r < 1; both are the same formula, but choosing well keeps the numbers positive. For example, 1 + 3 + 9 + ... to 8 terms = (38 − 1)/(3 − 1) = 6560/2 = 3280.
Worked example: Which term of the GP 2, 2√2, 4, ... is 128?Solution: a = 2 and r = 2√2/2 = √2. Set 2(√2)n−1 = 128, so (√2)n−1 = 64 = 26 = (√2)12. Hence n − 1 = 12 and 128 is the 13th term.
Sum of an infinite GP
If |r| < 1, then rn gets closer and closer to 0 as n grows, and Sn approaches a fixed value:
So 1 + 1/2 + 1/4 + ... = 1/(1 − 1/2) = 2. If |r| ≥ 1 the infinite series has no finite sum. A recurring decimal is a neat application: 0.333... = 3/10 + 3/100 + ... = (3/10)/(1 − 1/10) = 1/3.
Geometric mean and inserting GMs
The geometric mean of two positive numbers a and b is G = √(ab); then a, G, b form a GP. To insert n numbers between a and b so that the whole list is a GP, the common ratio is
and the inserted numbers are ar, ar², ..., arn. Inserting 3 GMs between 1 and 256 needs r4 = 256, so r = 4 and the means are 4, 16 and 64. (Taking r = −4 also satisfies r4 = 256; unless the question says otherwise, take the real positive ratio.)
Relation between AM and GM
For positive a and b, with A = (a + b)/2 and G = √(ab):
Also, if you know A and G, then a + b = 2A and ab = G², so a and b are the roots of x² − 2Ax + G² = 0.
Worked example: (a) The AM and GM of two positive numbers are 10 and 8. Find the numbers. (b) Find the minimum value of x + 9/x for x > 0.Solution: (a) a + b = 20 and ab = 64, so they are the roots of x² − 20x + 64 = 0, that is, (x − 16)(x − 4) = 0. The numbers are 16 and 4.
(b) By AM ≥ GM on the positive numbers x and 9/x: (x + 9/x)/2 ≥ √(x · 9/x) = 3, so x + 9/x ≥ 6. Equality holds when x = 9/x, that is, x = 3. The minimum is 6.
The method in (b) works whenever the product of the terms is a constant. Check that equality can actually happen; otherwise the bound is not the minimum.
Sums of special series
To sum a series whose nth term is a polynomial in n, expand an and apply these term by term. For example, if an = n(n + 1) = n² + n, then Sn = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(n + 2)/3.
Common mistakes: (1) Using a/(1 − r) when |r| ≥ 1. (2) Writing the nth term of a GP as arn; it is arn−1. (3) Using Sn = a(rn − 1)/(r − 1) when r = 1, where the formula breaks down. (4) Applying AM ≥ GM to numbers that are not all positive. (5) Taking the lower bound from AM ≥ GM as the minimum without checking that equality is possible. (6) Dividing by n instead of n + 1 when inserting n means.Exam focus
- nth term and sum of AP and GP, including finding which term equals a given value.
- Sum of an infinite GP and recurring decimals as fractions.
- Inserting arithmetic and geometric means between two numbers.
- AM ≥ GM for finding minimum or maximum values of expressions.
- Finding two numbers from their AM and GM.
- Sums of n, n² and n³ for series with a polynomial nth term.
Practice questions
The 20th term of the AP 3, 7, 11, ... is:
- 79
- 83
- 80
- 76
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The sum of the first 10 terms of the GP 1, 2, 4, 8, ... is:
- 1024
- 1023
- 2047
- 512
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Which term of the GP 2, 2√2, 4, ... is 128?
- 12th
- 14th
- 13th
- 7th
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The sum of the infinite GP 1 − 1/3 + 1/9 − 1/27 + ... is:
- 3/2
- 2/3
- 1/2
- 3/4
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The AM and GM of two positive numbers are 10 and 8. The numbers are:
- 12 and 8
- 16 and 4
- 18 and 2
- 15 and 5
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For x > 0, the minimum value of x + 9/x is:
- 9
- 3
- 6
- 10
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Three geometric means with positive common ratio are inserted between 1 and 256. The third of them is:
- 16
- 32
- 128
- 64
Show answer
The value of 1² + 2² + ... + 10² is:
- 385
- 55
- 3025
- 285





