Arithmetic and geometric progressions 55
Subtracting equation (1) from equation (2) gives
S n =
a(r n − 1)
(r − 1)
which is valid when r > 1.
For example, the sum of the first 8 terms of the GP 1, 2,
4, 8, 16, ... is given by S 8 =
1(2 8 − 1)
(2 − 1)
, since a = 1 and
r = 2
i.e. S 8 =
1(256 − 1)
1
= 255
Sum to infinity of a GP
When the common ratio r of a GP is less than unity, the
sum of n terms, S n =
a(1 −r n )
(1 −r)
, which may be written
as S n =
a
(1 −r)
−
ar n
(1 −r)
Since r < 1, r n becomes less as n increases, i.e. r n → 0
as n →∞.
Hence
ar n
(1 −r)
→ 0 as n →∞. Thus S n →
a
(1 −r)
as
n →∞.
The quantity
a
(1 −r)
is called the sum to infinity, S ∞ ,
and is the limiting value of the sum of an infinite number
of terms,
i.e. S ∞ =
a
(1 − r)
which is valid when −1
1 +
1
2 +
1
4 +· · · is
S ∞ =
1
1 −
1
2
, since a = 1 and r =
1
2 , 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 common ratio r of 2. The n’th term of a GP is ar n−1 ,
where a is 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,
1
2 , 1
1
2 , 4
1
2 , 13
1
2 , ...
1
2 , 1
1
2 , 4
1
2 , 13
1
2 , ... is a GP with a common ratio r = 3
The sum of n terms, S n =
a(r n − 1)
(r − 1)
Hence S 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 by ar 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 is ar 7 = (12)(1.4631719 ...) 7 = 172.3
The 11th term is ar 10 = (12)(1.4631719 ...) 10 = 539.7
Problem 14. Which term of the series 2187, 729,
243, ... is
1
9 ?
2187, 729, 243, ... is a GP with a common ratio r =
1
3
and first term a = 2187
The n’th term of a GP is given by: ar n−1
Hence
1
9
= (2187)
1
3
n−1
from which
1
3
n−1
=
1
(9)(2187)
=
1
3 2 3 7
=
1
3 9 =
1
3
9
Thus (n − 1) = 9, from which, n =9 + 1 =10
i.e.
1
9 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
72.0
= 0.8
also
ar 2
ar
=
46.08
57.6
= 0.8
Subtracting equation (1) from equation (2) gives
S n =
a(r n − 1)
(r − 1)
which is valid when r > 1.
For example, the sum of the first 8 terms of the GP 1, 2,
4, 8, 16, ... is given by S 8 =
1(2 8 − 1)
(2 − 1)
, since a = 1 and
r = 2
i.e. S 8 =
1(256 − 1)
1
= 255
Sum to infinity of a GP
When the common ratio r of a GP is less than unity, the
sum of n terms, S n =
a(1 −r n )
(1 −r)
, which may be written
as S n =
a
(1 −r)
−
ar n
(1 −r)
Since r < 1, r n becomes less as n increases, i.e. r n → 0
as n →∞.
Hence
ar n
(1 −r)
→ 0 as n →∞. Thus S n →
a
(1 −r)
as
n →∞.
The quantity
a
(1 −r)
is called the sum to infinity, S ∞ ,
and is the limiting value of the sum of an infinite number
of terms,
i.e. S ∞ =
a
(1 − r)
which is valid when −1
1 +
1
2 +
1
4 +· · · is
S ∞ =
1
1 −
1
2
, since a = 1 and r =
1
2 , 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 common ratio r of 2. The n’th term of a GP is ar n−1 ,
where a is 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,
1
2 , 1
1
2 , 4
1
2 , 13
1
2 , ...
1
2 , 1
1
2 , 4
1
2 , 13
1
2 , ... is a GP with a common ratio r = 3
The sum of n terms, S n =
a(r n − 1)
(r − 1)
Hence S 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 by ar 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 is ar 7 = (12)(1.4631719 ...) 7 = 172.3
The 11th term is ar 10 = (12)(1.4631719 ...) 10 = 539.7
Problem 14. Which term of the series 2187, 729,
243, ... is
1
9 ?
2187, 729, 243, ... is a GP with a common ratio r =
1
3
and first term a = 2187
The n’th term of a GP is given by: ar n−1
Hence
1
9
= (2187)
1
3
n−1
from which
1
3
n−1
=
1
(9)(2187)
=
1
3 2 3 7
=
1
3 9 =
1
3
9
Thus (n − 1) = 9, from which, n =9 + 1 =10
i.e.
1
9 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
72.0
= 0.8
also
ar 2
ar
=
46.08
57.6
= 0.8