geometric sequences and series part iii. geometric sequences and series the sequence is an example...

Geometric Sequences Geometric Sequences and Seriesand Series

Part IIIPart III

Geometric Sequences and Series

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.

With a calculator we can see that the sum is 186.But 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

TRICK 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



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 EX

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

Should use the factor theorem:

factor a is )1(04411)1( rf

We need to solve the cubic equation

04423 rrr


We will do this soon !!



factors the are )2)(2)(1( rrr


The solution to this cubic equation is therefore

04423 rrr

0)2)(2)(1( rrr

Since we were told we get

1r 2r



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


Sum to Infinity

IF |r|<1 then

)(1

))1(1(

r

aS

n

r

aS

1

Because (<1)∞ = 0

0


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 sequences and series part iii. geometric sequences and series the sequence is an example...

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