Arithmetic and geometric progressions 55

Subtracting equation (1) from equation (2) gives

Sn=

a(rn−1)

(r−1)

which is valid whenr> 1.

For example, the sum of the first 8 terms of the GP 1, 2,

4, 8, 16,...is given byS 8 =

1 ( 28 − 1 )

( 2 − 1 )

,sincea=1and

r= 2

i.e. S 8 =

1 ( 256 − 1 )

1

= 255

Sum to infinity of a GP

When the common ratiorof a GP is less than unity, the

sum ofnterms,Sn=

a( 1 −rn)

( 1 −r)

, which may be written

asSn=

a

( 1 −r)

−

arn
( 1 −r)
Sincer<1,rnbecomes less asnincreases, i.e.rn→ 0
asn→∞.

Hence

arn

( 1 −r)

→0asn→∞. ThusSn→

a

( 1 −r)

as

n→∞.
The quantity

a

( 1 −r)

is called thesum to infinity,S∞,

and is thelimitingvalueof thesum of an infinitenumber
of terms,

i.e.S∞=

a

(1−r)

which is valid when− 1 <r<1.

For example, the sum to infinity of the GP
1 +^12 +^14 +···is

S∞=

1

1 −^12

,sincea=1andr=^12 ,i.e.S∞= 2.

6.5 Worked problems on geometric

progressions

Problem 11. Determine the tenth term of the

series 3, 6, 12, 24,...

3, 6, 12, 24,...is a geometric progression with a com-
mon ratiorof 2. Then’th term of a GP isarn−^1 ,
where ais the first term. Hence the 10th term is:
( 3 )( 2 )^10 −^1 =( 3 )( 2 )^9 = 3 ( 512 )= 1536.

Problem 12. Find the sum of the first 7 terms of

the series,^12 ,1^12 ,4^12 ,13^12 ,...

1

2 ,1

1

2 ,4

1

2 ,13

1

2 ,...is a GP with a common ratior=^3

The sum ofnterms,Sn=

a(rn− 1 )

(r− 1 )

HenceS 7 =

1

2 (^3

(^7) − 1 )
( 3 − 1 )

1
2 (^2187 −^1 )
2
= 546
1
2
Problem 13. The first term of a geometric
progression is 12 and the fifth term is 55. Determine
the 8’th term and the 11’th term.
The 5th term is given byar^4 =55, where the first term
a= 12
Hence r^4 =
55
a

55
12
and r=^4
√(
55
12
)
= 1. 4631719 ...
The 8th term isar^7 =( 12 )( 1. 4631719 ...)^7 =172.3
The 11th term isar^10 =( 12 )( 1. 4631719 ...)^10 =539.7
Problem 14. Which term of the series 2187, 729,
243 ,...is^19?
2187, 729, 243,...is a GP with a common ratior=^13
and first terma= 2187
Then’th term of a GP is given by:arn−^1
Hence
1
9
=( 2187 )
( 1
3
)n− 1
from which
(
1
3
)n− 1

1
( 9 )( 2187 )

1
3237

1
39

(
1
3
) 9
Thus(n− 1 )=9, from which,n= 9 + 1 = 10
i.e.^19 is the 10th term of the GP.
Problem 15. Find the sum of the first 9 terms of
the series 72.0, 57.6, 46. 08 ,...
The common ratio,r=
ar
a

57. 6


58. 
    

    0
    = 0. 8
    (
    also
    ar^2
    ar

    
    


59. 08


60. 6
    = 0. 8
    )