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+131=10-3 + \frac{13}{1} = 10

    • 22−2+1=2122 - 2 + 1 = 21

    • 32−21+5=3532 - \frac{2}{1} + 5 = 35

Limiting Values of Sequences

  • The nth term of a sequence is given by n+22n+1\frac{n+2}{2n+1}.

  • Objective: Determine the limiting value of the sequence as n→∞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 nn approaches infinity (n→∞n \rightarrow \infty), the term n+22n+1\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→∞n \rightarrow \infty for the following expressions:

    • 6n−52n+3=62=3\frac{6n - 5}{2n + 3} = \frac{6}{2} = 3

    • −2n5n=−25\frac{-2n}{5n} = -\frac{2}{5}

    • 5n2−32n2+9=52\frac{5n^2 - 3}{2n^2 + 9} = \frac{5}{2}

    • 8n−25n−1=85\frac{8n - 2}{5n - 1} = \frac{8}{5}

Core Mini-Whiteboard Exercises

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

    • 3n2+18n2−1=38\frac{3n^2 + 1}{8n^2 - 1} = \frac{3}{8}

    • −8n+222n+11=−82=−4\frac{-8n + 22}{2n + 11} = -\frac{8}{2} = -4

    • 43(32)2=43⋅94=3\frac{4}{3} \left( \frac{3}{2} \right)^2 = \frac{4}{3} \cdot \frac{9}{4} = 3

Example 2: Limiting Values

  • lim⁡n→∞38n+13=0\lim_{n \to \infty} \frac{3}{8n + 13} = 0

  • 9=1(8+13)9 = 1(8 + 13)

  • 9=8+13h9 = 8 + 13h

  • n=13n = \frac{1}{3}

Example 3: Consecutive Terms Problems

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

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

  • Difference between consecutive terms:

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

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

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

    • 3n2+7n+4−(3n2+7n+2)(n+1)(n+2)\frac{3n^2 + 7n + 4 - (3n^2 + 7n + 2)}{(n+1)(n+2)}

    • 2(n+1)(n+2)\frac{2}{(n+1)(n+2)}

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

Core Mini-Whiteboard Exercises

  • The nth term of a sequence is 4n−10n\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 n2+n2\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)−(n2+n)=32(n+1)^2 + (n+1) - (n^2 + n) = 32

    • n2+2n+1+n+1−n2−n=32n^2 + 2n + 1 + n + 1 - n^2 - n = 32

    • 2n+2=322n + 2 = 32

    • 2n=302n = 30

    • n=15n = 15

    • 152+15=24015^2 + 15 = 240

    • 162+16=27216^2 + 16 = 272

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

  • 4n−10+4n−6n=8n−16n=8(n−2)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→∞n \rightarrow \infty:

    • (a) 3n9n−4\frac{3n}{9n - 4}

    • (b) n2n−7\frac{n}{2n - 7}

    • (c) 4n22n2+1\frac{4n^2}{2n^2 + 1}

    • (d) 8n+12n2−3\frac{8n + 12}{n^2 - 3}

    • (e) 3n+26n−4\frac{3n + 2}{6n - 4}

    • (f) 2n−915n−4\frac{2n - 9}{15n - 4}

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

    • (a) 22n2−1\frac{2}{2n^2 - 1}

    • (b) 4n22n2+1\frac{4n^2}{2n^2 + 1}

    • (c) 5n2+110n2−2\frac{5n^2 + 1}{10n^2 - 2}

    • (d) 2n2−5n2\frac{2n^2 - 5}{n^2}

  • (a) A sequence with nth term an+55n−1\frac{an + 5}{5n - 1} has a limiting value of 25\frac{2}{5} as n→∞n \rightarrow \infty. Work out the value of a.

  • (b) A sequence with nth term 10n−a3n+2\frac{10n - a}{3n + 2} has a limiting value of −32\frac{-3}{2} as n→∞n \rightarrow \infty. Work out the value of a.

Further Questions

  • (a) A sequence starts 23,35,47,59,…\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→∞n \rightarrow \infty.

  • (b) A sequence starts 15,49,713,1017,…\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→∞n \rightarrow \infty.

  • The nth term of a sequence is 240−8n70+4n\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→∞n \rightarrow \infty.

  • The nth term of a sequence is 3nn+5\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→∞n \rightarrow \infty.

  • The nth term of a sequence is n2+2n+1n^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−35n - 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 n+bn+c\frac{n + b}{n + c}.

  • The nth term of sequence Y is n+cn+b\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=2cb = 2c.

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

Linear Sequence Problems

  • A linear sequence starts a+b,a+3b,a+5b,a+7ba + 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=4a = -1, b = 4

Additional Problems

  • The first two terms in a linear sequence are 5+365 + 3\sqrt{6} and 6\sqrt{6}. What is the fifth term in the sequence?

  • The first term of a sequence is 5−2a5 - 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 n+22n−3\frac{n + 2}{2n - 3}. The nth term of sequence B is 3n−14n+5\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→∞n \rightarrow \infty:

    • (a) 3n9n−4\frac{3n}{9n - 4}

    • (b) n2n−7\frac{n}{2n - 7}

    • (c) 4n22n2+1\frac{4n^2}{2n^2 + 1}

    • (d) 8n+12n2−3\frac{8n + 12}{n^2 - 3}

    • (e) 3n+26n−4\frac{3n + 2}{6n - 4}

    • (f) 2n−915n−4\frac{2n - 9}{15n - 4}

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

    • (a) 22n2−1\frac{2}{2n^2 - 1}

    • (b) 4n22n2+1\frac{4n^2}{2n^2 + 1}

    • (c) 5n2+110n2−2\frac{5n^2 + 1}{10n^2 - 2}

    • (d) 2n2−5n2\frac{2n^2 - 5}{n^2}

  • (a) A sequence with nth term an+55n−1\frac{an + 5}{5n - 1} has a limiting value of 25\frac{2}{5} as n→∞n \rightarrow \infty. Work out the value of a.

  • (b) A sequence with nth term 10n−a3n+2\frac{10n - a}{3n + 2} has a limiting value of −32\frac{-3}{2} as n→∞n \rightarrow \infty. Work out the value of a.

  • (a) A sequence starts 23,35,47,59,…\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→∞n \rightarrow \infty.

  • (b) A sequence starts 15,49,713,1017,…\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→∞n \rightarrow \infty.

  • The nth term of a sequence is 240−8n70+4n\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→∞n \rightarrow \infty.

  • The nth term of a sequence is 3nn+5\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→∞n \rightarrow \infty.

  • The nth term of a sequence is n2+2n+1n^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−35n - 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: 912=828191^2 = 8281

    • 49×169=828149 \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.