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Geometric Sequences & Series

Learn how to solve problems involving growth and decay using geometric sequences and series. Topics include interest rates, population changes, and financial investments.

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Geometric Sequences & Series

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  1. Geometric Sequences & Series This week the focus is on using geometric sequences and series to solve problems involving growth and decay.

  2. Geometric Sequences & Series CONTENTS Growth and Decay Interest Rates Interest Rates Illustration Example 1 Example 2 Example 3 Assignment

  3. Geometric Sequences & Series Growth and Decay Geometric sequences can be used to solve problems involving growth or decay at a constant rate. Most commonly geometric sequences and series are used in problems involving interest rates and population growth/decay.

  4. Geometric Sequences & Series Growth and Decay Interest Rates: If property values are increasing at a rate of 4% per annum, the sequence of values increases as a geometric sequences with a common ratio of 1.04. Similarly, if a property decreased by 4% per annum the common ratio for the geometric sequence would be 0.96. The same principal also applies to interest rates for financial investments and population changes.

  5. Geometric Sequences & Series Growth and Decay Illustration: In 2000 a property was valued at £100,000. Over the next 3 years the property increased in value at a rate of 4% per annum. This gives the following values of the property: 2000 £100,000 2001 £104,000 2002 £108,160 2003 £112,486.40 a = 100,000 r = 1.04

  6. Geometric Sequences & Series Example 1: I invest £A in the bank at a rate of interest of 3.5% per annum. How long will it be before I double my investment? Solution: We can take the growth of the investment as a geometric sequence, the first term is A and the common ratio is 1.035 because the interest rate is 3.5%. The question is really asking what term will give us twice the amount we started with – i.e 2A. This means we are trying to find the value of n which gives 2A. continued on next slide

  7. Geometric Sequences & Series Solution continued: Substituting values into the general formula for the nth term of a geometric sequence gives: A(1.035)n = 2A (1.035)n = 2 by dividing across by A Take the log of both sides to solve the equation involving powers to get: log(1.035)n = log 2 n log(1.035) = log 2 n = log 2/log 1.035 = 20.15 Therefore my money will have doubled after 20.15 years.

  8. Geometric Sequences & Series Example 2: A car depreciates in value by 15% a year. If it is worth £11,054.25 after 3 years, what was its new price and when will it be first worth less than £5000. Solution: We can take the depreciation of the car as a geometric sequence with first term a (the new price) and the common ratio is 0.85 (100% - 15%). We know that after 3 years the value is £11,054.25. Although this value is after 3 years, this is the 4th term of the sequence. New Price a After 1 year ar After 2 years ar2 After 3 years ar3continued on next slide

  9. Geometric Sequences & Series Solution continued: Therefore ar3 = £11,054.25 Substitute in our value of r to get: a(0.85)3 = £11,054.25 a = £11,054.25/(0.85)3 = £18,000 Therefore the new price of the car was £18,000. For our geometric sequences we have a = 18,000 and r = 0.85 With these two values we can then find the value of n which gives a value less than 5000. continued on next slide

  10. Geometric Sequences & Series Solution continued: Solve for arn = 5000 18000(0.85)n = 5000 substitute in for a and r (0.85)n = 5000/18000 (0.85)n = 5/18 log (0.85)n = log (5/18) take log of both sides n log (0.85) = log (5/18) bring power forward n = log (5/18) / log (0.85) = 7.88 evaluate Therefore the value of the car will fall below £5000 after 7.88 years.

  11. Geometric Sequences & Series Example 3: In the sequence, 2, 6, 18, 54, ... Which is the first term to exceed 1 000 000? Solution: From the given sequence a = 2 and r = 3. If we want to find the first term to exceed 1 000 000 we solve arn = 1 000 000 Substitute in the values we have to get: 2(3)n = 1 000 000 (3)n = 500,000 dividing across by 2 continued on next slide

  12. Geometric Sequences & Series Solution continued: (3)n = 500,000 log (3)n = log (500,000) take log of both sides n log (3) = log (500,000) bring power forward n = log (500,000) / log (3) take numbers to one side n = 11.94 evaluate Therefore the first term to exceed 1 000 000 is the 12th term as we cannot have an 11.94th term.

  13. Geometric Sequences & Series ASSIGNMENT: This weeks assignment is in Moodle. Please follow the link in the course area. Attempt all questions and include working out. Deadline is 5:00pm on Monday 22nd March 2010.

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