Geometric and harmonic sequences – Sequences (Senior 5 Mathematics) | Ganzaa

Sequences

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Senior 5 · Mathematics

Sequences

4 lessons

Lesson 3 of 4

Geometric and harmonic sequences

25 min

Multiply instead of add

A bacterium that divides into two every 20 minutes gives $1, 2, 4, 8, 16, \dots$ cells: each number is twice the one before. After 24 hours, that is 72 divisions, there would be $2^{72} \approx 4.7 \times 10^{21}$ cells if nothing stopped the growth. Repeated multiplication by a fixed number makes the geometric sequences of this lesson. They model growth and decay, interest, and loans. The lesson ends with a third kind, the harmonic sequence, which is built from an arithmetic one.

Definition

Geometric sequence (geometric progression, GP): a sequence of non-zero numbers in which each term is the previous term multiplied by a fixed number $r$ :

u_{n+1} = r\,u_n \quad \text{for every } n.

Common ratio $r$ : the fixed quotient $\frac{u_{n+1}}{u_n}$ ( $r \ne 0$ ).

Example: $81, -27, 9, -3, \dots$ is geometric with $r = -\frac{1}{3}$ ; $1, 2, 4, 7, \dots$ is not, since $\frac{4}{2} \ne \frac{7}{4}$ .

General term and geometric means

From $u_{1}$ to $u_{n}$ there are $n - 1$ multiplications by $r$ :

u_n = u_1\,r^{n-1}, \qquad \text{and more generally} \qquad u_n = u_p\,r^{n-p}.

If $a, b, c$ are consecutive terms then $\frac{b}{a} = \frac{c}{b}$ , so $b^{2} = a c$ . For positive terms, $b = \sqrt{ac}$ : the middle term is the geometric mean of its neighbours. To insert $k$ geometric means between $a$ and $b$ , solve $b = a\,r^{k+1}$ for $r$ .

Figure 1. A geometric sequence with

0 \lt r \lt 1

V_n = 800 \times 0.8^n

thousand Frw (illustrative figures). The drop each year is 20% of a smaller and smaller value.

Definition

Geometric mean of two positive numbers $a$ and $c$ : the number $\sqrt{ac}$ .

Example: the geometric mean of 4 and 9 is $\sqrt{36} = 6$ , and $4, 6, 9$ is geometric with $r = \frac{3}{2}$ .

Worked example 1: two terms fix the sequence

In a geometric sequence $u_{2} = 12$ and $u_{5} = 324$ . Find $r$ , $u_{1}$ and $u_{n}$ .

Solution $u_5 = u_2\,r^3$ , so $r^3 = \frac{324}{12} = 27$ and $r = 3$ . Then $u_1 = \frac{12}{3} = 4$ , and $u_n = 4 \times 3^{n-1}$ .

Worked example 2: geometric means, two answers

Insert three geometric means between 2 and 162.

Solution Five terms, so $162 = 2 r^{4}$ , $r^{4} = 81$ and $r = 3$ or $r = - 3$ . The means are $6, 18, 54$ (with $r = 3$ ) or $- 6, 18, - 54$ (with $r = - 3$ ). Both are correct.

The sum of the first $n$ terms

Multiply $S_{n}$ by $r$ and subtract; all the middle terms cancel:

\begin{aligned} S_n &= u_1 + u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1} \\ rS_n &= \phantom{u_1 + {}} u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1} + u_1 r^n \\ S_n - rS_n &= u_1 - u_1 r^n. \end{aligned}

Important

Sum of a geometric sequence

S_n = u_1\,\frac{1 - r^n}{1 - r} = u_1\,\frac{r^n - 1}{r - 1} \quad (r \ne 1), \qquad S_n = n\,u_1 \quad (r = 1).

Use the second form when $r \gt 1$ , so that the top and bottom are both positive.

Worked example 3: sums

(a) Find the sum of the first 10 terms of $3 + 6 + 12 + \cdots$ . (b) How many terms of $2 + 6 + 18 + \cdots$ are needed for the sum to exceed 10,000?

Solution

$u_{1} = 3$ , $r = 2$ : $S_{10} = 3 \times \frac{2^{10} - 1}{2 - 1} = 3 \times 1023 = 3069$ .
$S_n = 2 \times \frac{3^n - 1}{3 - 1} = 3^n - 1$ . We need $3^n \gt 10\,001$ . Since $3^{8} = 6561$ and $3^9 = 19\,683$ , the answer is 9 terms. (Unit 3 solves such equations with logarithms.)

Worked example 4: depreciation

A school buys a laptop for 800,000 Frw. Its value falls by 20% each year (illustrative figures). Find its value after 4 years.

Solution Losing 20% leaves 80%, so each year's value is 0.8 times the last: $V_n = 800\,000 \times 0.8^n$ . After 4 years, $V_4 = 800\,000 \times 0.4096 = 327\,680$ Frw (Figure 1).

Harmonic sequences

The numbers $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$ are neither arithmetic nor geometric, but their reciprocals $1, 2, 3, 4, \dots$ are arithmetic. Sequences of this kind appear wherever a quantity is shared equally, such as time per job or average speed.

Definition

Harmonic sequence (harmonic progression, HP): a sequence of non-zero numbers whose reciprocals $\frac{1}{u_1}, \frac{1}{u_2}, \frac{1}{u_3}, \dots$ form an arithmetic sequence.

Harmonic mean of two non-zero numbers $a$ and $b$ (with $a + b \ne 0$ ): the number $H$ with $\frac{1}{H} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right)$ , that is

H = \frac{2ab}{a + b}.

Example: $6, 4, 3, \frac{12}{5}, 2$ is harmonic, because $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{5}{12}, \frac{1}{2}$ go up by $\frac{1}{12}$ each time.

Worked example 5: harmonic means

Insert two harmonic means between 3 and 12.

Solution Work with the reciprocals: insert two arithmetic means between $\frac{1}{3}$ and $\frac{1}{12}$ . With four terms, $d = \frac{\frac{1}{12} - \frac{1}{3}}{3} = -\frac{1}{12}$ , giving $\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}$ . Turning back: $3, 4, 6, 12$ . The harmonic means are 4 and 6.

Worked example 6: average speed

A bus travels from one town to another at $60\,\mathrm{km\,h^{-1}}$ and returns along the same road at $40\,\mathrm{km\,h^{-1}}$ . Find its average speed for the round trip.

Solution Let the distance one way be $s$ km. The times are $\frac{s}{60}$ h and $\frac{s}{40}$ h, so

\bar{v} = \frac{2s}{\frac{s}{60} + \frac{s}{40}} = \frac{2}{\frac{1}{60} + \frac{1}{40}} = \frac{2 \times 60 \times 40}{60 + 40} = 48\,\mathrm{km\,h^{-1}},

the harmonic mean of 60 and 40, not their arithmetic mean 50. The slower leg takes longer, so it counts more.

Three means compared

For positive numbers $a$ and $b$ , let $A = \frac{a + b}{2}$ , $G = \sqrt{ab}$ and $H = \frac{2ab}{a + b}$ . Then $A - G = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0$ , and $AH = \frac{a + b}{2}\cdot\frac{2ab}{a + b} = ab = G^2$ , so $H = \frac{G^2}{A} \le G$ . Hence

A \ge G \ge H, \quad \text{with equality only when } a = b.

Figure 2. Three means of

a = 4

and

b = 1

in one drawing, to scale. The radius

OD = \frac{a + b}{2}

; the half-chord

CD = \sqrt{ab}

(the altitude theorem from S3); and

DE = \frac{CD^2}{OD} = \frac{2ab}{a + b}

(the leg theorem in triangle OCD). Since a hypotenuse is longer than a leg,

OD \ge CD \ge DE

Warning

Signs and the ratio. $r^{4} = 81$ has two real solutions, $r = 3$ and $r = - 3$ ; do not drop the negative one. And a geometric sequence can never contain 0: if $u_1 r^{n-1} = 0$ then $u_{1} = 0$ or $r = 0$ , both excluded.

Tip

Which kind is it? Constant differences: arithmetic. Constant ratios: geometric. Reciprocals with constant differences: harmonic. Test with three consecutive terms $a, b, c$ : $2 b = a + c$ , $b^{2} = a c$ or $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ .

Khan Academy: Introduction to geometric sequences (about 11 min). Watch how the common ratio is found by dividing a term by the one before, and the closing problem about a bungee jump that rebounds to a fixed fraction of the previous height.

✦

Practice

The last two are harder.

Find the 8th term of $5, -10, 20, \dots$ .
In a geometric sequence $u_{3} = 18$ and $u_{6} = - 486$ . Find $r$ and $u_{1}$ .
Find the geometric mean of 4 and 25.
Find the sum of the first 8 terms of $1 + \frac{1}{2} + \frac{1}{4} + \cdots$ .
Find the sum $5 + 15 + 45 + \cdots + 3645$ .
Show that $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}$ are consecutive terms of a harmonic sequence, and give the next term.
Insert two harmonic means between 2 and 8.
For $a = 9$ and $b = 4$ , find $A$ , $G$ and $H$ , and check that $G^{2} = A H$ .

Show answersHide answers

$r = - 2$ : $u_{8} = 5 (- 2)^{7} = - 640$ .
$r^3 = \frac{-486}{18} = -27$ , $r = - 3$ , $u_1 = \frac{18}{9} = 2$ .
$\sqrt{100} = 10$ .
$S_8 = \frac{1 - \left(\frac{1}{2}\right)^8}{1 - \frac{1}{2}} = 2\left(1 - \frac{1}{256}\right) = \frac{255}{128}$ .
$3645 = 5 \times 3^{n-1}$ , so $3^{n-1} = 729 = 3^6$ and $n = 7$ . $S_7 = 5 \times \frac{3^7 - 1}{2} = 5 \times 1093 = 5465$ .
The reciprocals $3, 5, 7$ go up by 2. Next reciprocal 9, so the next term is $\frac{1}{9}$ .
Reciprocals $\frac{1}{2}$ to $\frac{1}{8}$ in 3 steps: $d = -\frac{1}{8}$ , giving $\frac{1}{2}, \frac{3}{8}, \frac{1}{4}, \frac{1}{8}$ . The harmonic means are $\frac{8}{3}$ and 4.
$A = 6.5$ , $G = 6$ , $H = \frac{72}{13} \approx 5.54$ . $AH = 6.5 \times \frac{72}{13} = 36 = G^2$ . ✓

Sources and credits

Written by Ganzaa: the explanations, the graph, the construction of the three means, the worked examples (the laptop figures are illustrative), the practice items and the quiz questions.
Video: Introduction to geometric sequences, Khan Academy, embedded from YouTube.

1Multiply instead of add

2General term and geometric means

3The sum of the first n n n terms

4Harmonic sequences

5Three means compared

Multiply instead of add

General term and geometric means

The sum of the first $n$ terms

Harmonic sequences

Three means compared