Sequence and Series – SS1 Further Mathematics Lesson Note (First Term)

A sequence is a set of numbers arranged in a definite order, where each term follows a rule.

Behavioural objectives

1. Define a sequence, a term and a series.
2. Write down terms of a sequence from its general term T_n.
3. Identify an A.P, its first term (a) and its common difference (d).
4. Use T_n = a + (n – 1)d to find a term, the common difference or the number of terms.
5. Insert arithmetic means between two given numbers.
6. Find the sum of the first n terms of an A.P.

What are a sequence and a series?

A sequence is a pattern of numbers arranged in a particular order, and a series is the sum of the terms of a sequence.

• Sequence: numbers listed in order, related by a well-defined rule.
• Term: each number in the sequence.
• T₁, T₂, T₃: the first, second and third terms.
• Series: the terms of the sequence added together.

Example

Consider the sequence 1, 4, 7, 10, 13 …

• T₁: 1 is the first term.
• T₂: 4 is the second term.
• T₃: 7 is the third term.
• Series: 1 + 4 + 7 + 10 + 13 = 35.

What is the difference between a sequence and a series?

A sequence is a list of numbers in order. A series is what you get when you add those numbers.

How do you find the nth term of a sequence?

The nth term, T_n, is a formula that gives any term of a sequence from its position n.

Finding T_n for 5, 9, 13, 17, 21 …

• T₁ = 5 + 4(0)
• T₂ = 5 + 4(1)
• T₃ = 5 + 4(2)
• T₄ = 5 + 4(3)
• T_n = 5 + 4(n – 1) = 5 + 4n – 4 = 4n + 1
• When n = 30, T₃₀ = 4(30) + 1 = 121.

Worked examples: terms from a general term

Example 1. Write down the first four terms of the sequence with T_n = (n + 1)/(3n + 2).

1. T₁ = (1 + 1)/(3(1) + 2) = 2/5
2. T₂ = (2 + 1)/(3(2) + 2) = 3/8
3. T₃ = (3 + 1)/(3(3) + 2) = 4/11
4. T₄ = (4 + 1)/(3(4) + 2) = 5/14

Example 2. Write down the first four terms of the sequence with T_n = 5 × (1/2)^n-2.

1. T₁ = 5 × (1/2)^-1 = 5 × 2 = 10
2. T₂ = 5 × (1/2)⁰ = 5 × 1 = 5
3. T₃ = 5 × (1/2)¹ = 5/2
4. T₄ = 5 × (1/2)² = 5/4
5. The sequence is 10, 5, 5/2, 5/4 …

What is an arithmetic progression (A.P)?

An arithmetic progression (A.P), also called a linear sequence, is a sequence in which each term is found by adding a constant to the term before it.

• First term (a): the first number in the A.P.
• Common difference (d): the constant added each time, found by subtracting any term from the next one.
• Increasing A.P: d is positive, as in 2, 4, 6, 8 where d = 2.
• Decreasing A.P: d is negative, as in 10, 6, 2, -2 where d = -4.
• n: the position of a term, or the number of terms.

What is the formula for the nth term of an A.P?

The nth term of an A.P is T_n = a + (n – 1)d.

How the formula comes about

• T₁ = a
• T₂ = a + d
• T₃ = a + 2d
• T₄ = a + 3d
• T₅ = a + 4d

For any A.P, T_n = U_n = a + (n – 1)d, where:

• T_n or U_n: the nth term.
• a: the first term.
• d: the common difference.
• n: the position of the term.

Worked examples

Example 1. What is the 10th term of the sequence 10, 6, 2, -2 …?

1. a = 10, d = 6 – 10 = -4, n = 10.
2. T₁₀ = 10 + (10 – 1)(-4)
3. T₁₀ = 10 + 9(-4) = 10 – 36
4. T₁₀ = -26

Example 2. Find the term of the A.P 3½, 7, 10½ … which is 77.

1. a = 3½, d = 7 – 3½ = 3½, T_n = 77, n = ?
2. 77 = 3½ + (n – 1)3½
3. 77 = 3½ + 3½n – 3½
4. 77 = 3½n
5. n = 77 ÷ 7/2 = 77 × 2/7 = 22
6. 77 is the 22nd term.

Example 3. The first term of an A.P is 3 and the 8th term is 31. Find the common difference.

1. a = 3, T₈ = 31, n = 8, d = ?
2. 31 = 3 + (8 – 1)d
3. 31 – 3 = 7d
4. d = 28/7 = 4

What is an arithmetic mean?

An arithmetic mean is a term placed between two numbers so that the numbers form an A.P.

• Middle term of three: if a, b, c are consecutive terms of an A.P, b is the arithmetic mean of a and c.

Why b = ½(a + c)

1. The common difference is the same, so b – a = c – b.
2. Then b + b = a + c.
3. So 2b = a + c.
4. Therefore b = ½(a + c).

How do you insert arithmetic means between two numbers?

1. Treat the first number as a and the last number as the last term.
2. If k means are inserted, the number of terms is n = k + 2.
3. Use T_n = a + (n – 1)d to find d.
4. Add d repeatedly to a to get each mean.

Example. Insert four arithmetic means between -5 and 10.

1. The sequence is -5, p, q, r, s, 10, so a = -5, n = 6 and T₆ = 10.
2. 10 = -5 + (6 – 1)d
3. 15 = 5d, so d = 3.
4. p = -5 + 3 = -2, q = -2 + 3 = 1, r = 1 + 3 = 4, s = 4 + 3 = 7.
5. The numbers are -5, -2, 1, 4, 7, 10.

How do you find a and d when two terms are given?

Write each given term using a + (n – 1)d, then solve the two equations together.

Example. The 8th term of a linear sequence is 18 and the 12th term is 26. Find the first term, the common difference and the 20th term.

1. T₈ = a + 7d = 18 … (i)
2. T₁₂ = a + 11d = 26 … (ii)
3. Subtract (i) from (ii): 4d = 8, so d = 2.
4. Put d = 2 in (i): a + 14 = 18, so a = 4.
5. T₂₀ = a + 19d = 4 + 19 × 2 = 42.

How do you find the sum of terms in an A.P?

The sum of the first n terms of an A.P is S_n = n/2 (2a + (n – 1)d).

How the formula is obtained

1. Write the sum forwards: S_n = a + (a + d) + (a + 2d) + … + T_n … (i)
2. Write it backwards: S_n = T_n + (T_n – d) + (T_n – 2d) + … + a … (ii)
3. Add (i) and (ii). Every pair gives (a + T_n), and there are n pairs.
4. So 2S_n = n(a + T_n), and S_n = n/2 (a + T_n).
5. Replace T_n with a + (n – 1)d to get S_n = n/2 (2a + (n – 1)d).

Which formula should you use?

• S_n = n/2 (2a + (n – 1)d): use when a and d are known.
• S_n = n/2 (a + T_n): use when the first and last terms are known.

Worked examples

Example 1. Find the sum of the first 25 terms of the A.P 3, 10, 17 …

1. a = 3, d = 7, n = 25.
2. S₂₅ = 25/2 (2 × 3 + (25 – 1)7)
3. S₂₅ = 25/2 (6 + 168) = 25/2 × 174
4. S₂₅ = 2175

Example 2. Find the sum of the first eight terms of a linear sequence whose first term is 6 and last term is 46.

1. a = 6, T₈ = 46, n = 8.
2. S₈ = 8/2 (6 + 46)
3. S₈ = 4 × 52 = 208

Example 3. The sum of the first ten terms of an A.P is 255 and the sum of the first twenty terms is 1010. Find the sum of the next twenty terms after the first ten.

1. S₁₀ = 5(2a + 9d) = 255, so 2a + 9d = 51 … (i)
2. S₂₀ = 10(2a + 19d) = 1010, so 2a + 19d = 101 … (ii)
3. Subtract (i) from (ii): 10d = 50, so d = 5.
4. Put d = 5 in (i): 2a + 45 = 51, so a = 3.
5. The next twenty terms are terms 11 to 30, so the sum is S₃₀ – S₁₀.
6. S₃₀ = 15(6 + 29 × 5) = 15 × 151 = 2265
7. S₃₀ – S₁₀ = 2265 – 255 = 2010

Evaluation

1. State the difference between a sequence and a series.
2. Write down the first three terms of the sequence whose general term is T_n = 2n + 3.
3. Find the 12th term of the A.P 7, 11, 15 …
4. Insert two arithmetic means between 4 and 19.
5. Find the sum of the first 10 terms of the A.P 7, 11, 15 …

Assignment

Answer all three parts.

1. The sum of the first four terms of a linear sequence (A.P) is 26 and that of the next four terms is 74. Find (i) the first term and (ii) the common difference.
2. Calculate (i) the common difference and (ii) the 20th term of the A.P 100, 96, 92, 88, 84 …
3. The first three terms of an A.P are x, 2x + 1 and 4x + 1. Find x and the sum of the first 18 terms.