geometric series

32: Geometric 32: Geometric Sequences and SeriesSequences and Series

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© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

Geometric Sequences and Series

Module C2

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There is a legend that Zarathustra, a Persian, invented chess to give interest to the life of the king who was bored. For his reward, Zarathustra asked for a quantity of grain, according to the following rules.


2 on the next,

1 grain was to be placed on the 1st square of the chess board,

4 on the 3rd

and so on, doubling the number each square.

How many must be placed on the 64th square?


...,8,4,2,1

We have a sequence:

Each term is twice the previous term, so by the 64th term we have multiplied by 2 sixty-three timesWe have 63

64 2u

approximately or 9 followed by 18 zeros!

18109

Geometric Sequence


632...,8,4,2,1

The sequence

is an example of aGeometric sequence

A sequence is geometric if

rterm previous

term each

where r is a constant called the common

ratio

In the above sequence, r = 2


A geometric sequence or geometric progression (G.P.) is of the form

The nth term of an G.P. is

1 nn aru

...,,,, 32 ararara


Exercises1. Use the formula for the nth term to find the

term indicated of the following geometric sequences

term th6...,32,8,2

term th5...,4

3,3,12

term th7...,0020,020,2.0

(b)

(c)

(a)

Ans: 2048)4(2 5

Ans: 64

3

4

112

4

Ans: 00000020)1.0(20 6 .


e.g.1 Evaluate

Writing out the terms helps us to recognize the G.P.

5

1

)2(3n

n

5432 )2(3)2(3)2(3)2(3)2(3

Summing terms of a G.P.

Although with a calculator we can see that the sum is 186, we need a formula that can be used for any G.P.

The formula will be proved next but you don’t need to learn the proof.


4325 ararararaS

Subtracting the expressions gives

With 5 terms of the general G.P., we have

Multiply by r: 5432

5 arararararrS

Move the lower row 1 place to the right

43255 arararararSS

5432 ararararar





Multiply by r:

and subtract

54325 arararararrS

5432 ararararar

43255 arararararSS

4325 ararararaS



5432 ararararar



Multiply by r:

555 ararSS

4325 ararararaS

54325 arararararrS

43255 arararararSS



r

raS

1

)1( 5

5

r

raS

n

n

1

)1(

Similarly, for n terms we

get

555 ararSS So,

Take out the common factors

and divide by ( 1 – r )

)1()1( 5rr aS5



gives a negative denominator if r

> 1

r

raS

n

n

1

)1(The formula

1

)1(

r

raS

n

n

Instead, we can use


To get this version of the formula, we’ve multiplied the 1st form by

1

1


5432 )2(3)2(3)2(3)2(3)2(3 For our series

12

)12(6 5

nS

1

)31(6

186

52,6 nra and

1

)1(

r

raS

n

nUsing



Find the sum of the first 20 terms of the geometric series, leaving your answer in index form

31

312 20

20

Sr

raS

n

n

1

)1(

...541862 e.g. 2

2

6,2

raSolution

:

3

1

3

We’ll simplify this answer without using a calculator



4

312 20

2

31 20

There are 20 minus signs here and 1 more outside the bracket!

31

312 20

20

S

1

2



e.g. 3In a geometric sequence, the sum of the 3rd and 4th terms is 4 times the sum of the 1st and 2nd terms. Given that the common ratio is not –1, find its possible values.

Solution: As there are so few terms, we don’t need the formula for a sum 3rd term + 4th term = 4( 1st term + 2nd

term ))(432 araarar

Divide by a since the 1st term, a, cannot be zero:

)1(432 rrr 04423 rrr



factor anot is )1(04411)1( rf

Using the factor theorem:

0))(1( r 2r

factor a is )1(04411)1( rf

4

We need to solve the cubic equation

04423 rrr

The squared term in is 2r04423 rrr

As we already have , the quadratic factor is complete. The middle term is zero.

2r



factor anot is )1(04411)1( rf

Using the factor theorem:

0))(1( r 2r

factor a is )1(04411)1( rf

4

We need to solve the cubic equation

04423 rrr

0)2)(2)(1( rrrSince we were told we

get 1r 2r

42 rFactorizing



e.g. 4 £100 is invested every year on the first of January and earns compound interest at the rate of 4% per annum. Find the amount by the end of the 5th year, to the nearest penny.Solution: The last £100 is invested for 1 year

only.

041100 The 4th £100 is invested for 2 years so at the end is worth

2041100

At the end, this £100 is worth

100040100 )0401(100

100 is a common factor



1

)1(

r

raS

n

n

1041

1041104 5

5

S

305635 £S (nearest penny)

At the end of the 5 years, the total invested will be worth

52 041100...041100041100

This is a G.P. with 104041100 a

5,041 nr


SUMMARY

r

raS

n

n

1

)1(

A geometric sequence or geometric progression (G.P.) is of the form

The nth term of an G.P. is

1 nn aru

...,,,, 32 ararara

The sum of n terms is

1

)1(

r

raS

n

no

r


Exercises

1. Find the sum of the first 15 terms of the following G.P., giving the answers in index form

2 + 8 + 32 + . . .

2. Find the sum of the first 15 terms of the G.P.

4 2 + 1 + . . . giving your answer

correct to 3 significant figures.


Exercises

3

)14(2 15

15

S

1

)1(

r

raS

n

n15,4,2 nra

14

)14(2 15

15

S

1. Solution: 2 + 8 + 32 + . . .

501

5014 15

15

S

r

raS

n

n

1

)1(15,50,4 nra

2. Solution: 4 2 + 1 + . . .

67215 S( 3 s.f. )

geometric series

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