MEI C2 Study Resources Core2 Sequences And Series 3

MEI Core 2 Sequences and Series Section 3: Geometric sequences and series Study Plan Background Geometric series are very important and appear in many areas of everyday life (not least in how bank accounts operate!). Again you will need to be very familiar with the notation and formulae. Detailed work plan 1. Read through the section writing down any formulae that you find so that you can refer to them easily. Example 7.9 shows methods of solutions to problems that you cannot yet solve accurately by using a calculator and/or a spreadsheet. The techniques used here are transferable to other areas of study e.g. business or science. Example 7.11 shows how the sum to infinity of a geometric series can be useful for finding the fractional equivalent of a recurring decimal. 2. To see additional examples, look at the Flash resources nth term of a GP, Sum of a GP and Sum of an infinite GP. There is also a Geometric series spreadsheet in which you can vary the values of the first term and the common difference. 3. Exercise 7C Try all of the questions labelled M.E.I. plus questions 2, 3, 5, 7, 9, 11, 14, 18. Questions 7*, 9*, 12*, 15*, 18* have worked solutions that you can download. 4. For extra practice try the interactive questions Finding terms in a geometric series, Finding the sum of a geometric series, Finding the first term of a finite GP, Finding the number of terms in a finite GP, Number of terms in a GP to exceed a given sum, Finding the common ratio in an infinite GP, and Finding the first term of an infinite GP.

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MEI Core 2 Sequences and series Section 3: Geometric sequences and series Notes and Examples These notes contain subsections on  Formulae for geometric sequences  Worked examples  Finding the number of terms

Formulae for geometric sequences In the section on Arithmetic sequences, you used two basic formulae to solve problems: the formula for the kth term in an arithmetic sequence, and the formula for the sum of the first n terms of the sequence. When you are working with geometric sequences, you need to use the equivalent formulae, and you may also need to use a third formula, for the sum to infinity of the sequence. The formula for the kth term of a geometric sequence is ak  ar k 1 where a is the first term of the sequence and r is the common ratio (the number that each term is multiplied by to obtain the next term).

The formula for the sum of the first n terms of a geometric sequence is Sn 

a (1  r n ) 1 r

or

Sn 

a (r n  1) r 1

These two formulae are equivalent. If r is less than 1, then it is easier to use the left-hand formula, and if r is greater than 1, it is easier to use the righthand version. However, either will give you the right answer! The formula for the sum to infinity of a geometric sequence is

S 

a 1 r

for –1 < r < 1

The restriction –1 < r < 1 is very important, as it is only for these values of r that the series converges (i.e. the terms become numerically smaller and smaller). If r >1, the terms become larger and larger and so the sum of the terms becomes larger and larger, so it makes no sense to talk about the sum to infinity. If r < -1, the terms become numerically larger and larger, but alternate between positive and negative, so the sum of the terms becomes

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MEI C2 Sequences Section 3 Notes and Examples alternately large and small, and again it makes no sense to talk about the sum to infinity. In both the cases r > 1 and r < -1, the series diverges.

You can explore geometric series using the interactive Geometric series spreadsheet. Try varying the values of a and r. Notice in particular how the terms oscillate when r is negative.

Worked examples Example 1 A geometric sequence has first term 4 and common ratio ½. (i) Find the 6th term. (ii) Find the sum of the first 10 terms. (iii) Find the sum to infinity of the terms of the sequence. Solution (i) ak  ar k 1

Substituting a = 4, r = ½ and k = 6

6th term = 4  12  0.125 5

a (1  r n ) 1 r 10 4(1  12 ) 10 S10   8(1  12 )  7.992 1 1 2

(ii) S n 

a 1 r 4  1  12

(iii) S 

Substituting a = 4, r = ½ and n = 10

Substituting a = 4, and r = ½

8 You can see additional examples using the Flash resources nth terms of a GP, Sum of a GP, and Sum of an infinite GP.

For practice in questions like Example 1(i), try the interactive resource Finding terms in a geometric series. For practice in questions like Example 1(iii), try the interactive resource Finding the sum of a geometric series.

Example 2 A geometric series has common ratio 0.5 and the sum of the first 5 terms is 77.5. Find the first term.

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MEI C2 Sequences Section 3 Notes and Examples Solution a(1  r n ) Sn  1 r a 1  0.55  77.5  1  0.5

77.5 

Substituting r = 0.5, Sn = 77.5 and n = 5

0.96875a 0.5

a  40 The first term is 40.

For practice in questions like Example 2, try the interactive resource Finding the first term of a finite G.P.

In Example 3, the common ratio is negative. Be careful when you are using your calculator to work out powers of a negative number – you will probably need to use brackets. If you type in –2^8 then the calculator will probably work out 28 and then apply a negative. So you need to type in (-2)^8. Remember that even powers of a negative number are positive, and odd powers are negative, and check that your answer is sensible.

Example 3 A geometric sequence has 3rd term 12 and 6th term –96. (i) Find the first term and the common ratio. (ii) Find the sum of the first 8 terms. Solution  ar 2  12 (i) 3rd term = ar 2  ar 5  96 6th term = ar 5 Dividing the second equation by the first gives r 3  8 r = –2 2 ar  12  4a  12  a  3 The first term is 3 and the common ratio is –2.

a (r n  1) r 1 3((2)8  1) S8  2  1 3(256  1)  3  255

(ii) S n 

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MEI C2 Sequences Section 3 Notes and Examples In the next example, you have to use the formula for the sum of an infinite geometric series.

Example 4 (i) The sum of an infinite geometric series with common ratio -0.25 is 6.4. Find the first term. (ii) The sum of an infinite geometric series with first term 3 is 5. Find the common ratio. Solution (i)

S 

a 1 r

6.4 

a 1   0.25 

a 1.25 a  6.4  1.25  8 The first term is 8. 

(ii)

a 1 r 3 5 1 r 3 1  r   0.6 5 r  1  0.6  0.4 The common ratio is 0.4. S 

For practice in questions like Example 4(i), try the interactive resource Finding the first term of an infinite G.P. For practice in questions like Example 4(ii), try the interactive resource Finding the common ratio in an infinite G.P.

Finding the number of terms In Examples 5 and 6, you have to find the value of n, which appears as an index. If you have covered logarithms, you can use these to find n. If not, you can still solve the problem, by using trial and improvement. In Example 5, it is quite simple to use trial and improvement. In Example 6, both methods are shown.

Example 5 The sum of a geometric series with first term 3 and common ratio -2 is 513. Find the number of terms in the series.

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MEI C2 Sequences Section 3 Notes and Examples Solution a (1  r n ) Sn  1 r 3 1  ( 2) n  513  1  (2)  513 

3 1  ( 2) n 

3 513  1  ( 2) n ( 2) n  512

By trial and improvement n = 9. There are 9 terms in the series.

For practice in questions like the one above, try the interactive resource Finding the number of terms in a finite G.P.

In the next example you have to find the number of terms required to exceed a given sum.

Example 6 Tom saves money every year. The first year he saves £100. Each year he increases the amount he saves by 10%. After how many years do Tom’s savings first exceed £1000 (excluding any interest he has earned)? Solution The amount Tom saves each year forms a geometric sequence with a = 100 and r = 1.1. The sum of Tom’s savings after n years is given by the formula 100(1.1n  1) Sn  1.1  1 100(1.1n  1)  0.1  1000(1.1n  1) When this amount exceeds 1000 1000(1.1n  1)  1000 1.1n  1  1 1.1n  2

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MEI C2 Sequences Section 3 Notes and Examples …or by trial and improvement

Solve the inequality either using logarithms…

Using logarithms 1.1n  2

Using trial and improvement 1.17  1.95 1.18  2.14

log1.1n  log 2 n log1.1  log 2 log 2 n log1.1 n  7.27 The savings first exceed £1000 after 8 years.

For practice in questions like the one above, try the interactive resource Number of terms in a G.P to exceed a given sum.

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MEI Core 2 Sequences and series Section 3: Geometric sequences and series Crucial points 1.

Know the formulae Make sure that you know the formulae for the kth term of a geometric sequence, the sum of the first n terms of a geometric sequence, and the sum to infinity of a convergent geometric sequence. It’s a good idea to make sure you can prove these formulae. This will help you to remember them. See the Notes and Examples.

2.

Beware of simple arithmetical errors Be careful if the common ratio of a geometric sequence is negative. Make sure that you are using your calculator correctly to find powers of a negative number – use brackets if necessary.

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Core 2 Sequences and Series Section 3: Geometric sequences and series Exercise 1. Find the 10th term of the geometric sequence 20, 16, 12.8, … 2. The first term of a geometric sequence is 1 and the common ratio is 3. (i) Find the 7th term. (ii) Find the sum of the first 8 terms. 3. A geometric sequence has first term 2 and common ratio 0.75. (i) Find the 4th term. (ii) Find the sum of the first 5 terms. (iii) Find the sum to infinity of the sequence. 4. A geometric sequence has 3rd term 18 and 6th term -60.75. Find the first term and the common ratio. 5. A geometric series has first term 2 and sum to infinity 5. (i) Find the common ratio. (ii) Find the sum of the first 10 terms of the series. (iii) How many terms are needed for the sum of the series to exceed 4.99? 6. The numbers 5x + 1, 4x – 4 and 3x – 5 form three consecutive terms of a geometric sequence. (i) Find the two possible values of x. (ii) Find the common ratio corresponding to each possible value of x.

  = 0.454545... can be written as the infinite 7. The infinite recurring decimal 0.45 geometric series 0.45 + 0.0045 + 0.000045 + … (i) Write down the first term and the common ratio of this geometric series.   as an exact (ii) Find the sum to infinity of the series and hence express 0.45 fraction in its lowest terms.    as 8. Using the same method as in Question 7, express the recurring decimal 0.407 an exact fraction in its lowest terms. 9. Kelly works for the same company for 10 years. Her starting salary is £18000, and each year she receives a pay rise of 4%. (i) How much does Kelly earn in the 10th year? (ii) How much has she earned in total over the 10 year period?

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C2 Series Section 3 Exercise 10. A ball is dropped from a height of 2 metres. After each bounce it rebounds to a height 0.8 times the height that it reached after the last bounce. (i) After how many bounces does the ball first rebound to less than 10 cm from the ground? (ii) Find the total distance travelled by the ball before it comes to rest. (iii) After how many bounces has the ball travelled more than 99% of the total distance it travels before coming to rest?

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Core 2 Sequences and Series Section 3: Geometric sequences and series Solutions to Exercise 1. First term a = 20

16 = 0.8 20 = 20 × 0.8 10 = 2.68 (3 s.f.)

Common ratio r = 10th term = ar

9

2. First term a = 1 Common ratio r = 3 (i) 7th term = ar 6 = 1 × 36 = 729 (ii) S 8 =

a(r 8 − 1) 1(38 − 1) = = 3280 r −1 3−1

3. First term a = 2 Common ratio r = 0.75 (i) 4th term = ar 3 = 2 × 0.75 3 = 0.84375

a(1 − r 5 ) 2(1 − 0.75 5 ) = = 6.10 (3 s.f.) (ii) S 5 = 1 −r 1 − 0.75 (iii) S ∞ =

a 1 −r

=

2 =8 1 − 0.75

⇒ ar 2 = 18 (1) 5 ⇒ ar = −60.75 (2) −60.75 = −3.375 r3 = Dividing (2) by (1): 18 r = −1.5 Substituting into (1): a( −1.5 )2 = 18 a =8 The first term is 8 and the common ratio is -1.5.

4. 3rd term = 18 6th term = -60.75

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C2 Series Section 3 Exercise solutions a

S∞ =

5. (i)

1 −r 2 5= 1 −r 1 − r = 0.4

r = 0.6 (ii) S 10 =

(iii)

2(1 − r 10 ) 2(1 − 0.6 10 ) = = 4.97 (3 s.f.) 1 −r 1 − 0.6

2(1 − 0.6n ) > 4.99 1 − 0.6 1 − 0.6n > 0.998

0.6n < 0.002 Since the log of a number less than 1 is negative, the inequality needs to be reversed

log0.6n < log0.002

n log0.6 < log0.002 log0.002 log0.6 n > 12.2 13 terms are needed.

n>

6. (i) Common ratio =

4x − 4 3x − 5 and 4x − 4 5x +1

4x − 4 3x − 5 = 5 x + 1 4x − 4 (4 x − 4)2 = (3 x − 5 )(5 x + 1)

16 x 2 − 32 x + 16 = 15 x 2 − 22 x − 5

x 2 − 10 x + 21 = 0 ( x − 3)( x − 7) = 0 x = 3 or x = 7 (ii) Common ratio =

4x − 4 5x +1

12 − 4 = 0.5 15 + 1 28 − 4 2 For x = 7, common ratio = =3 35 + 1

For x = 3, common ratio =

7. (i) First term a = 0.45 Common ratio r = 0.01

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C2 Series Section 3 Exercise solutions (ii) S ∞ =

a 1 −r

=

0.45 0.45 45 5 = = = 1 − 0.01 0.99 99 11

   = 0.407 + 0.000407 + 0.000000407 + .... 8. 0.407 First term a = 0.407 Common ratio r = 0.001 a 0.407 0.407 407 11 = = = = S∞ = 1 − r 1 − 0.001 0.999 999 27

9. Annual salary is a geometric sequence First term a = 18000 Common ratio r = 1.04 (i) 10th term = ar (ii) S 10 =

9

= 18000 × 1.04 9 = £25619.61 (to nearest penny)

a(r 10 − 1) 18000(1.0410 − 1) = = £216107.93 (to nearest penny) r −1 1.04 − 1

10. Heights rebounded to after each bounce form a geometric sequence After 1 bounce, it rebounds to 2 × 0.8 = 1.6 metres First term a = 1.6 Common ratio r = 0.8 (i) nth term = ar n − 1 = 1.6 × 0.8 n − 1 1.6 × 0.8 n − 1 < 0.1 0.8 n − 1 < 0.0625 log0.8 n − 1 < log0.0625 (n − 1)log0.8 n − 1 < log0.0625 log0.0625 log0.8 n − 1 > 12.4

n −1 >

n > 13.4

It first rebounds to less than 10 cm after 14 bounces. (ii) Total distance travelled = 2 + 2 S ∞ = 2 +2×

a

1 −r 2 × 1.6 =2+ 1 − 0.8 = 18 metres

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C2 Series Section 3 Exercise solutions (iii) Total distance travelled after n bounces = 2 + 2 Sn 1.6(1 − 0.8 n ) 2 +2× > 0.99 × 18 1 − 0.8 2 + 16(1 − 0.8 n ) > 17.82 1 − 0.8 n > 0.98875 0.8 n < 0.01125

n log0.8 < log0.01125 log0.01125 log0.8 n > 20.1 After 21 bounces.

n>

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MEI Core 2 Sequences and series Section 3: Geometric sequences and series Multiple Choice Test Questions 1 to 3 refer to the geometric sequence 2, 6, 18, 54 … 1) The 8th term of the sequence is (a) 4374 (c) 279936 (e) I don’t know

(b) 13122 (d) 2187

2) The sum of the first 10 terms of the sequence is (a) 19683 (c) 59048 (e) I don’t know

(b) 118096 (d) 39366

3) The first term which is greater than 100000 is (a) the 10th term (c) the 9th term (e) I don’t know

(b) the 11th term (d) the 12th term

Questions 4 to 6 refer to the geometric sequence 24, -12, 6, -3 … 4) The 7th term of the sequence is (a) 0.1875 (c) 0.375 (e) I don’t know

(b) 0.75 (d) 0.09375

5) The sum to infinity of the sequence is (a) 2 (c) 12 (e) I don’t know

(b) 48 (d) 16

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MEI C2 Sequences Section 3 MC test solutions 6) After how many terms is the sum of the sequence within 1% of the sum to infinity of the sequence? (a) 6 terms (c) 8 terms (e) I don’t know

(b) 7 terms (d) 10 terms

7) A geometric sequence has 3rd term 36 and 5th term 81. Find the first term, a, and the common ratio, r. (a) a = 24, r = 32 (c) a = 16, r = 23 (e) I don’t know

(b) a = 16, r = (d) a = 24, r =

3 2 2 3

8) A geometric sequence has first term 5 and sum to infinity 6.25. The common ratio of the sequence is (a) 0.2 (c) 0.25 (e) I don’t know

(b) 0.8 (d) 1.25

Anna’s grandmother gives her £15 for her 10th birthday. She tells Anna that she will increase the amount she gives her for her birthday by £5 each year until she is 18. Amy’s grandmother gives her £15 for her 10th birthday. She tells Amy that she will increase the amount she gives her for her birthday by 25% each year until she is 18.

9) On which birthday is Amy first given more than Anna? (a) 13th (c) 15th (e) I don’t know

(b) 16th (d) 14th

10) How much more money has Amy received in total than Anna by the time the girls are 18? (Give your answer to the nearest pound). (a) £40 (c) £18 (e) I don’t know

(b) £72 (d) £24

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MEI Core 2 Sequences and series Chapter Assessment 1. An arithmetic series has first term 3 and common difference 5. Find (i) the 4th term (ii) the sum of the first 12 terms.

[2] [3]

2. The 5th term of an arithmetic series is 16 and the 10th term is 30. (i) Find the first term and the common difference. [4] (ii) How many terms of the series are needed for the sum of the series to exceed 1000? [4] 3. A geometric series has first term 2 and common ratio 0.2. Find (i) the 3rd term (ii) the sum of the first 4 terms of the series (iii) the sum to infinity of the series.

[2] [3] [2]

4. A geometric series has 1st term 3 and sum to infinity 8. Find the common ratio.

[4]

5. A geometric series has first term 54 and 4th term 2. (i) What is the common ratio? [3] (ii) Find the sum to infinity of the series. [2] (iii) After how many terms is the sum of the series greater than 99% of the sum to infinity? [5] 6. When Sarah is 5 years old, her parents start to give her pocket money of 50p per week. On her birthday each year, her parents increase her pocket money by 50p. (i) How much pocket money does Sarah get in the first year? [2] (ii) How much more money in total does Sarah get in the second year than the first year? [3] (iii) How much money has Sarah been given in total by her 11th birthday? [3] (iv) After how many complete years is the total amount Sarah has been given more than £1000? [4] 7. At the beginning of each month, Mark puts £N from his salary into a savings account. At the end of every month, interest is added to his savings at the rate of r% per month. (i) Write down an expression for the amount of money in Mark’s account at the end of (a) 1 month (b) 2 months (c) 3 months, and hence show that the amount of money in Mark’s account at the end of n months is given by 2 3 n r N (1 + 100 [8] ) + N (1 + 100r ) + N (1 + 100r ) + .... + N (1 + 100r )

(ii) Use the formula for a geometric progression to simplify this expression. [3] (iii) How much does Mark have after 5 years if he saves £100 a month at an interest rate of 0.5% per month? [3] Total 60 marks

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