Algebra - Sequences

Sequences and Limiting Values

Starter Problems: Finding the nth Term

• Find the nth term of the following sequences:

  • a) 1, 5, 13, 25, …

  • b) 10, 7, 4, 1, …

  • e) 7, 23, 49, 85, …

• Find the nth term of the sequence with a first term of 10 and a common ratio of 4.

• Find the nth term of the sequence with a first term of 250 and a common ratio of 0.2.

• Find the nth term of the sequence with a first term of 8 and a common ratio of -5.

• Examples:

  • -3 + \frac{13}{1} = 10

  • 22 - 2 + 1 = 21

  • 32 - \frac{2}{1} + 5 = 35

Limiting Values of Sequences

• The nth term of a sequence is given by \frac{n+2}{2n+1}.

• Objective: Determine the limiting value of the sequence as n \rightarrow \infty.

• Consider the values of the sequence for increasing values of n:

  • n = 10: 0.571429…

  • n = 100: 0.507436…

  • n = 1000: 0.500750…

  • n = 10000: 0.500075…

  • n = 100000: 0.500008…

  • n = 1000000: 0.500001…

• As n approaches infinity (n \rightarrow \infty), the term \frac{n+2}{2n+1} approaches 0.5.

• Therefore, the limiting value of the sequence is 0.5.

Example 1: Determining Limiting Values

• Determine the limiting value of the sequence as n \rightarrow \infty for the following expressions:

  • \frac{6n - 5}{2n + 3} = \frac{6}{2} = 3

  • \frac{-2n}{5n} = -\frac{2}{5}

  • \frac{5n^2 - 3}{2n^2 + 9} = \frac{5}{2}

  • \frac{8n - 2}{5n - 1} = \frac{8}{5}

Core Mini-Whiteboard Exercises

• Find the limiting value of the sequence as n \rightarrow \infty for the expressions:

  • \frac{3n^2 + 1}{8n^2 - 1} = \frac{3}{8}

  • \frac{-8n + 22}{2n + 11} = -\frac{8}{2} = -4

  • \frac{4}{3} \left( \frac{3}{2} \right)^2 = \frac{4}{3} \cdot \frac{9}{4} = 3

Example 2: Limiting Values

• \lim_{n \to \infty} \frac{3}{8n + 13} = 0

• 9 = 1(8 + 13)

• 9 = 8 + 13h

• n = \frac{1}{3}

Example 3: Consecutive Terms Problems

• A sequence has an nth term defined as \frac{3n}{n+1}.

• Objective: Show that the difference between two consecutive terms is \frac{1}{(n+1)(n+2)}.

• Difference between consecutive terms:

  • \frac{3(n+1) + 1}{(n+1) + 1} - \frac{3n + 1}{n + 1}

  • \frac{3n + 4}{n+2} - \frac{3n + 1}{n+1}

  • \frac{(3n + 4)(n+1) - (3n + 1)(n+2)}{(n+1)(n+2)}

  • \frac{3n^2 + 7n + 4 - (3n^2 + 7n + 2)}{(n+1)(n+2)}

  • \frac{2}{(n+1)(n+2)}

• The difference between two consecutive terms is indeed \frac{2}{(n+1)(n+2)}.

Core Mini-Whiteboard Exercises

• The nth term of a sequence is \frac{4n - 10}{n}. Prove that the sum of any two consecutive terms of the sequence is divisible by 8.

• The nth term of a sequence is \frac{n^2 + n}{2}. Two consecutive terms in the sequence have a difference of 32. Work out the two terms.

  • (n+1)^2 + (n+1) - (n^2 + n) = 32

  • n^2 + 2n + 1 + n + 1 - n^2 - n = 32

  • 2n + 2 = 32

  • 2n = 30

  • n = 15

  • 15^2 + 15 = 240

  • 16^2 + 16 = 272

• \frac{4(n+1) - 10}{n} + \frac{4n - 6}{n}

• \frac{4n - 10 + 4n - 6}{n} = \frac{8n - 16}{n} = \frac{8(n-2)}{n}

• Therefore, the sum of any two consecutive terms of the sequence is divisible by 8.

Questions and Exercises

• Find the limits of the following sequences as n \rightarrow \infty:

  • (a) \frac{3n}{9n - 4}

  • (b) \frac{n}{2n - 7}

  • (c) \frac{4n^2}{2n^2 + 1}

  • (d) \frac{8n + 12}{n^2 - 3}

  • (e) \frac{3n + 2}{6n - 4}

  • (f) \frac{2n - 9}{15n - 4}

• Find the limits of the following sequences as n \rightarrow \infty:

  • (a) \frac{2}{2n^2 - 1}

  • (b) \frac{4n^2}{2n^2 + 1}

  • (c) \frac{5n^2 + 1}{10n^2 - 2}

  • (d) \frac{2n^2 - 5}{n^2}

• (a) A sequence with nth term \frac{an + 5}{5n - 1} has a limiting value of \frac{2}{5} as n \rightarrow \infty. Work out the value of a.

• (b) A sequence with nth term \frac{10n - a}{3n + 2} has a limiting value of \frac{-3}{2} as n \rightarrow \infty. Work out the value of a.

Further Questions

• (a) A sequence starts \frac{2}{3}, \frac{3}{5}, \frac{4}{7}, \frac{5}{9}, …. Find the nth term for this sequence and the limiting value as n \rightarrow \infty.

• (b) A sequence starts \frac{1}{5}, \frac{4}{9}, \frac{7}{13}, \frac{10}{17}, …. Find the nth term for this sequence and the limiting value as n \rightarrow \infty.

• The nth term of a sequence is \frac{240 - 8n}{70 + 4n}.

  • (a) Work out the term in the sequence that is equal to 0.

  • (b) Write down the limiting value of the sequence as n \rightarrow \infty.

• The nth term of a sequence is \frac{3n}{n + 5}.

  • (i) Explain why 1 is not a term in this sequence.

  • (ii) Work out the limiting value of the sequence as n \rightarrow \infty.

• The nth term of a sequence is n^2 + 2n + 1. Prove that the sum of any two consecutive terms of the sequence is an odd number.

• The nth term of the linear sequence 2, 7, 12, 17, … is 5n - 3. A new sequence is formed by squaring each term of the linear sequence and adding 1. Prove algebraically that all the terms in the new sequence are multiples of 5.

• (a) Write down the nth term of the linear sequence 4, 7, 10, 13, …

  • $3n + 1$

• (b) Hence, write down the nth term of the quadratic sequence 16, 49, 100, 169, …

  • $(3n + 1)^2 = 9n^2 + 6n + 1$

• (c) For the sequence in (b), show that the 30th term is equal to the product of the 2nd and 4th terms.

  • 2nd term: 49

  • 4th term: 169

  • 30th term: $91^2 = 8281$

  • $49 \times 169 = 8281$

Sequence X and Sequence Y

• The nth term of sequence X is \frac{n + b}{n + c}.

• The nth term of sequence Y is \frac{n + c}{n + b}.

  • (a) Show that the sequences have the same first term.

  • (b) The 2nd term of sequence X is equal to the 3rd term of sequence Y. Show that b = 2c.

  • (c) Prove that: \frac{b}{c} = \frac{2c + 1}{c + 2}.

Linear Sequence Problems

• A linear sequence starts a + b, a + 3b, a + 5b, a + 7b. The 5th and 8th terms have values 35 and 59.

  • (a) Work out a and b.

  • (b) Work out the nth term of the sequence.

  • a = -1, b = 4

Additional Problems

• The first two terms in a linear sequence are 5 + 3\sqrt{6} and \sqrt{6}. What is the fifth term in the sequence?

• The first term of a sequence is 5 - 2a. The term-to-term rule of the sequence is subtract 4a and then multiply by 2. The fourth term of the sequence is 58. Work out the second term of the sequence.

• The nth term of sequence A is \frac{n + 2}{2n - 3}. The nth term of sequence B is \frac{3n - 14}{n + 5}. The qth term in sequence A is the same as the qth term in sequence B. Work out the value of q.

Further Exercises

• Find the limits of the following sequences as n \rightarrow \infty:

  • (a) \frac{3n}{9n - 4}

  • (b) \frac{n}{2n - 7}

  • (c) \frac{4n^2}{2n^2 + 1}

  • (d) \frac{8n + 12}{n^2 - 3}

  • (e) \frac{3n + 2}{6n - 4}

  • (f) \frac{2n - 9}{15n - 4}

• Find the limits of the following sequences as n \rightarrow \infty:

  • (a) \frac{2}{2n^2 - 1}

  • (b) \frac{4n^2}{2n^2 + 1}

  • (c) \frac{5n^2 + 1}{10n^2 - 2}

  • (d) \frac{2n^2 - 5}{n^2}

• (a) A sequence with nth term \frac{an + 5}{5n - 1} has a limiting value of \frac{2}{5} as n \rightarrow \infty. Work out the value of a.

• (b) A sequence with nth term \frac{10n - a}{3n + 2} has a limiting value of \frac{-3}{2} as n \rightarrow \infty. Work out the value of a.

• (a) A sequence starts \frac{2}{3}, \frac{3}{5}, \frac{4}{7}, \frac{5}{9}, …. Find the nth term for this sequence and the limiting value as n \rightarrow \infty.

• (b) A sequence starts \frac{1}{5}, \frac{4}{9}, \frac{7}{13}, \frac{10}{17}, …. Find the nth term for this sequence and the limiting value as n \rightarrow \infty.

• The nth term of a sequence is \frac{240 - 8n}{70 + 4n}.

  • (a) Work out the term in the sequence that is equal to 0.

  • (b) Write down the limiting value of the sequence as n \rightarrow \infty.

• The nth term of a sequence is \frac{3n}{n + 5}.

  • (i) Explain why 1 is not a term in this sequence.

  • (ii) Work out the limiting value of the sequence as n \rightarrow \infty.

• The nth term of a sequence is n^2 + 2n + 1. Prove that the sum of any two consecutive terms of the sequence is an odd number.

• The nth term of the linear sequence 2, 7, 12, 17, … is 5n - 3. A new sequence is formed by squaring each term of the linear sequence and adding 1. Prove algebraically that all the terms in the new sequence are multiples of 5.

• (a) Write down the nth term of the linear sequence 4, 7, 10, 13, …

  • $3n + 1$

• (b) Hence, write down the nth term of the quadratic sequence 16, 49, 100, 169, …

  • $(3n + 1)^2 = 9n^2 + 6n + 1$

• (c) For the sequence in (b), show that the 30th term is equal to the product of the 2nd and 4th terms.

  • 2nd term: 49

  • 4th term: 169

  • 30th term: 91^2 = 8281

  • 49 \times 169 = 8281

Review and Completion

• Complete the booklet. Check your answers against the marking scheme.

• Note what the marks are awarded for.

• Teams message any questions.