New
New
Year 11
•
Higher
Advanced problem solving with further sequences
I can use my enhanced knowledge of sequences to solve problems.
New
New
Year 11
•
Higher
Advanced problem solving with further sequences
I can use my enhanced knowledge of sequences to solve problems.
Lesson details
Key learning points
- Compound interest can be thought of as a geometric sequence.
- You can apply your knowledge of geometric sequences to solve algebraic compound interest problems.
Common misconception
A multiplier of 1.3 is used to increase a value by 3%
A multiplier of 1.3 is used to increase a value by 30%
Keywords
Interest - Interest is money added to savings or loans.
Compound interest - Compound interest is calculated on the original amount and the interest accumulated over the previous period.
Exponential - The general form for an exponential equation is y = ab^x where a is the coefficient, b is the base and x is the exponent.
The second learning cycle draws on pupils' understanding of proof as they consider proofs which involve quadratic sequences. Some pupils may benefit from reviewing proof before doing this part of the lesson.
Teacher tip
Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
Video
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Starter quiz
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6 Questions
Q1.
Interest which is calculated on the original amount and the interest accumulated over the previous period is called interest.
Correct Answer: compound
compound
Q2.
If a geometric sequence starts 75, 105, 147, ... what is the common ratio? Give your answer as a decimal.
Correct Answer: 1.4
1.4
Q3.
£100 is invested at 1% compound interest a month. Which calculation would find the value of the money after 5 months?
$$100 \times 5(1.01)$$
$$100 \times 5(1.1)$$
Correct answer: $$100 \times 1.01^5$$
$$100 \times 1.01^5$$
$$100 \times 1.1^5$$
Q4.
Which of these are examples of exponential equations?
$$y=x^2$$
$$y=100x^3$$
Correct answer: $$y=2^x$$
$$y=2^x$$
Correct answer: $$y = 3 \times 5^x$$
$$y = 3 \times 5^x$$
$$y= x^{1\over 2}$$
Q5.
Which of these is the general form for an odd number when $$n$$ is an integer?
$$n$$
$$n+1$$
$$2n$$
Correct answer: $$2n +1$$
$$2n +1$$
$$3n$$
Q6.
What is the $$n^\text{th}$$ term of the quadratic sequence which starts 1, 3, 9, 19, 33, ...?
$$n^2 -1$$
Correct answer: $$2n^2 -4n +3$$
$$2n^2 -4n +3$$
$$2n^2+4n-3$$
$$4n^2 -10n + 7$$
$$4n^2 -2n-1$$
Exit quiz
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6 Questions
Q1.
£1600 is borrowed with yearly compound interest and after a year £2240 is owed. What is the yearly interest rate?
1.4%
14%
25%
30%
Correct answer: 40%
40%
Q2.
A loans company charges 4% interest a month. If £35 152 was owed after three months, how much was borrowed?
Correct Answer: £31 250, 31 250, £31250, 31250
£31 250, 31 250, £31250, 31250
Q3.
£40 000 was borrowed with compound interest and after 4 years £58 564 was owed. What was the yearly interest rate? Assume no repayments were made.
Correct answer: 10%
10%
14%
37%
45%
50%
Q4.
£24 000 is borrowed at a compound interest rate of 5% a month. What would be the $$n^\text{th}$$ term rule for the geometric sequence formed from the amount owed each month starting with month 1?
$$24 000n + 1.05$$
$$1.05n + 24 000$$
$$24000 \times n^{1.05}$$
Correct answer: $$24 000 \times 1.05^n$$
$$24 000 \times 1.05^n$$
Q5.
Some money was borrowed with compound interest. After 2 years £4000 was owed and after 4 years £6250 was owed. After how many years would the loan value be greater than £10 000 ?
Correct Answer: 7, 7 years
7, 7 years
Q6.
Which of these is the $$n^\text{th}$$ term rule for the triangular numbers? (1, 3, 6, 10, 15, ...)
$$n^2 -n$$
Correct answer: $$\frac{n(n+1)}{2}$$
$$\frac{n(n+1)}{2}$$
$$\frac{n^2-(n-1)}{2}$$
$$\frac{1}{2}n^2 -\frac{1}{2}n + \frac{1}{2}$$
Correct answer: $$\frac{1}{2}n^2 +\frac{1}{2}n$$
$$\frac{1}{2}n^2 +\frac{1}{2}n$$