Sequence and Series

A sequence is a list of numbers in an order, made by a rule: its numbers are its terms, t₁, t₂, t₃, …, and a formula for $t_n$ is its general term. Written with plus signs between the terms, a sequence becomes a series. In an arithmetic sequence each term is the one before plus the same number, the common difference d, so $t_n = a + (n - 1)d$. In a geometric sequence each term is the one before times the same number, the common ratio r, so $t_n = ar^{n-1}$. In both, a is the first term.

Grade 9 asked one kind of question: what is a given term? This unit asks two new ones. What are the terms between two given terms (the means)? And what do the first n terms add up to (the sum)?

Activity 1: the number in the gap

The terms between the first and last terms of an arithmetic sequence are called arithmetic means (AMs). In 3, 7, 11, 15, 19 the three arithmetic means between 3 and 19 are 7, 11 and 15: the numbers that split the gap from 3 to 19 into equal steps (Figure 6.2).

One arithmetic mean. If a, m, b is an arithmetic sequence, m − a = b − m, so

$$m = \frac{a + b}{2}$$

The single arithmetic mean of two numbers is their average, the same “mean” you meet in statistics.

Several arithmetic means. For m arithmetic means between a and b, the sequence has the means plus the two ends, so n = m + 2 terms, and b is the last, the nth. So b = a + (n − 1)d gives the common difference, and the means are a + d, a + 2d, … up to a + md. In one line, the gap b − a is cut into m + 1 equal steps:

$$d = \frac{b - a}{m + 1}$$

Worked examples: arithmetic means

Activity 2: a monthly incentive

a. By Rs 200 each month: the incentives make an arithmetic sequence with a = 800, d = 200.

b. You can add the seven numbers. But look at them written forwards and backwards:

$$\begin{aligned} S_7 &= 800 + 1000 + 1200 + 1400 + 1600 + 1800 + 2000 \\ S_7 &= 2000 + 1800 + 1600 + 1400 + 1200 + 1000 + 800 \end{aligned}$$

Add the two lines column by column: every column makes 2,800, the first month plus the last.

$$2S_7 = 7 \times 2800, \quad\text{so}\quad S_7 = \frac{7}{2} \times 2800 = 9800$$

The total was Rs 9,800. Each pair adds to the same amount because, going along, one row rises by 200 exactly as fast as the other falls by 200.

c. Baisakh 2080 is the 13th month: $t_{13} = 800 + 12 \times 200 = 3200$. Rs 3,200.

d. Here the pairing trick works the same way without writing all 13 numbers: 13 pairs of (first + last), halved.

$$S_{13} = \frac{13}{2}(800 + 3200) = \frac{13}{2} \times 4000 = 26000$$

Rs 26,000 in 13 months.

The formula, and why it works

A series whose terms make an arithmetic sequence is an arithmetic series. Write its first n terms forwards and backwards, with a the first term and l (or $t_n$) the last. Each column adds to a + l, and there are n columns, so twice the sum is n(a + l):

$$S_n = \frac{n}{2}(a + l)$$

In words: the sum is the number of terms times the average of the first and last terms. Since l = a + (n − 1)d, the same sum without the last term is

$$S_n = \frac{n}{2}[2a + (n - 1)d]$$

Use the first when you know the last term, the second when you know the common difference.

Worked examples: sums

Question 1

Question 2

Find the arithmetic mean between:

Question 3

Find the arithmetic means:

Question 4

Each of these is an arithmetic sequence. Find x.

Question 5

Find the sum of each arithmetic series.

Question 6

The first term, the last term and the sum of all the terms of an arithmetic series are given. Find its common difference.

Question 7

The common difference, the number of terms and the sum of an arithmetic series are given. Find its first term.

Question 8

Question 9

Question 10

Question 11

Activity 1: the gap in a geometric sequence

The terms between the first and last terms of a geometric sequence are called geometric means (GMs). Arithmetic means split a gap into equal steps; geometric means split it into equal ratios (Figure 6.3).

One geometric mean. If a, m, b is a geometric sequence of positive numbers, m/a = b/m, so $m^2 = ab$ and

$$m = \sqrt{ab}$$

Several geometric means. For m geometric means between a and b there are again n = m + 2 terms, b is the nth, and $b = ar^{n-1}$ gives the common ratio. The means are ar, $ar^2$, … up to $ar^m$.

Worked examples: geometric means

Activity 2: a salary that rises 10% a month

Each month’s salary is the last one times 1.1:

Month	1	2	3	4	5
Salary (Rs)	10,000	11,000	12,100	13,310	14,641

They make a geometric sequence with a = 10000 and r = 1.1. Adding them gives the total, but here is a trick that works for any number of terms. Write the sum, then the sum times the ratio:

$$\begin{aligned} S_5 &= 10000 + 11000 + 12100 + 13310 + 14641 \quad \text{(i)} \\ 1.1\,S_5 &= 11000 + 12100 + 13310 + 14641 + 16105.1 \quad \text{(ii)} \end{aligned}$$

Multiplying by the ratio moves every term along one place: four of the five terms are in both lines. Subtract (i) from (ii) and they all cancel, leaving only the last term of (ii) and the first of (i):

$$\begin{aligned} (1.1 - 1)S_5 &= 16105.1 - 10000 \\ 0.1\,S_5 &= 10000(1.1^5 - 1) \\ S_5 &= \frac{10000(1.1^5 - 1)}{1.1 - 1} = 61051 \end{aligned}$$

He earned Rs 61,051 in 5 months.

The formula, and why it works

A series whose terms make a geometric sequence is a geometric series. With first term a, common ratio r and n terms,

$$\begin{aligned} S_n &= a + ar + ar^2 + \cdots + ar^{n-1} \\ rS_n &= ar + ar^2 + \cdots + ar^{n-1} + ar^n \end{aligned}$$

Subtracting, $rS_n - S_n = ar^n - a$, so $(r - 1)S_n = a(r^n - 1)$ and

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

When r is between −1 and 1 (a fraction such as $\frac12$), both brackets are negative, and it is tidier to write the same formula the other way up:

$$S_n = \frac{a(1 - r^n)}{1 - r}$$

If the last term $l = ar^{n-1}$ is known instead of n, use $ar^n = lr$:

$$S_n = \frac{lr - a}{r - 1}$$

Worked examples: geometric sums

Back to the basketball

By its tenth bounce the ball has traveled about 7.15 m, and however long it bounces it will never travel more than 7.2 m.

The ball falls 1.8 m. Then each of the next 9 bounces goes up and comes down the same height, 1.8 × 0.6 = 1.08 m, then 1.08 × 0.6, and so on: a geometric series with a = 1.08, r = 0.6, n = 9, counted twice.

$$\begin{aligned} \text{Distance} &= 1.8 + 2 \times \frac{1.08(1 - 0.6^9)}{1 - 0.6} \\ &= 1.8 + 2 \times 2.673 \\ &\approx 7.15 \text{ m} \end{aligned}$$

As the bounces go on, $0.6^9$ and the powers after it shrink towards 0, so the bracket never passes 1.08 ÷ 0.4 = 2.7: the whole journey stays under 1.8 + 2 × 2.7 = 7.2 m. Try it with other balls:

In the SEE’s Compulsory Mathematics paper, Algebra carries 15 marks in three questions, with knowledge, understanding, application and higher-ability parts spread among them (CDC’s specification grid). In CDC’s model paper the sequence question is about a pattern of savings:

Part	Level	Marks	What it asks
a.	knowledge	1	is the sequence arithmetic or geometric?
b.	understanding	2	the total after some terms
c.	higher ability	2	is the total enough by a certain time? with a reason

Here is a question in that shape, and one to practice with.

Question 1

Question 2

Find the geometric mean between:

Question 3

Find the geometric means:

Question 4

Each of these is a geometric sequence. Find x.

Question 5

Find the sum of each geometric series.

Question 6

The first term, the last term and the sum of a geometric series are given. Find the common ratio.

Question 7

Question 8

Question 9