Arithmetic and geometric progressions 57
(a) Let the GP be a, ar, ar
2
, ... , ar
n
The first term a = £100
The common ratio r = 1.08
Hence the second term is
ar = (100)(1.08) = £108,
which is the value after 1 year,
the third term is
ar
2
= (100)(1.08)
2
= £116.64,
which is the value after 2 years, and so on.
Thus the value after 10 years
= ar
10
= (100)(1.08)
10
= £215.89
(b) When £300 has been reached, 300 =ar n
i.e.
300 = 100(1.08)
n
and
3 = (1.08)
n
Taking logarithms to base 10 of both sides gives:
lg 3 = lg(1.08)
n
= n lg(1.08),
by the laws of logarithms
from which, n =
lg 3
lg1.08
= 14.3
Hence it will take 15 years to reach more than
£300.
Problem 20. A drilling machine is to have 6
speeds ranging from 50 rev/min to 750 rev/ min. If
the speeds form a geometric progression determine
their values, each correct to the nearest whole
number.
Let the GP of n terms be given by a, ar, ar 2 , ..., ar n−1 .
The first term a = 50 rev/min
The 6th term is given by ar 6−1 , which is 750 rev/min,
i.e.
ar
5
= 750
from which r
5
=
750
a
=
750
50
= 15
Thus the common ratio, r =
5
√
15 = 1.7188
The first term is a = 50 rev/min
the second term is ar = (50) (1.7188) = 85.94,
the third term is ar 2 = (50) (1.7188) 2 = 147.71,
the fourth term is ar 3 = (50) (1.7188) 3 = 253.89,
the fifth term is ar
4
= (50) (1.7188)
4
= 436.39,
the sixth term is ar 5 = (50) (1.7188) 5 = 750.06
Hence, correct to the nearest whole number, the 6 speeds
of the drilling machine are 50, 86, 148, 254, 436 and
750 rev/min.
Now try the following exercise
Exercise 27 Further problems on geometric
progressions
1. In a geometric progression the 5th term is
9 times the 3rd term and the sum of the 6th
and 7th terms is 1944. Determine (a) the common ratio, (b) the first term and (c) the sum
of the 4th to 10th terms inclusive.
[(a) 3 (b) 2 (c) 59022]
2. Which term of the series 3, 9, 27, ... is
59049?
[10 th ]
3. The value of a lathe originally valued at
£3000 depreciates 15% per annum. Calculate
its value after 4 years. The machine is sold
when its value is less than £550. After how
many years is the lathe sold?
[£1566, 11 years]
4. If the population of Great Britain is 55 million
and is decreasing at 2.4% per annum, what
will be the population in 5 years time?
[48.71 M]
5. 100 g of a radioactive substance disintegrates
at a rate of 3% per annum. How much of the
substance is left after 11 years?
[71.53 g]
6. If £250 is invested at compound interest of
6% per annum determine (a) the value after
15 years, (b) the time, correct to the nearest
year, it takes to reach £750.
[(a) £599.14 (b) 19 years]
7. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the
speeds form a geometric progression determine their values, each correct to the nearest
whole number.
[100, 139, 193, 268, 373, 518,
720, 1000 rev/min]
(a) Let the GP be a, ar, ar
2
, ... , ar
n
The first term a = £100
The common ratio r = 1.08
Hence the second term is
ar = (100)(1.08) = £108,
which is the value after 1 year,
the third term is
ar
2
= (100)(1.08)
2
= £116.64,
which is the value after 2 years, and so on.
Thus the value after 10 years
= ar
10
= (100)(1.08)
10
= £215.89
(b) When £300 has been reached, 300 =ar n
i.e.
300 = 100(1.08)
n
and
3 = (1.08)
n
Taking logarithms to base 10 of both sides gives:
lg 3 = lg(1.08)
n
= n lg(1.08),
by the laws of logarithms
from which, n =
lg 3
lg1.08
= 14.3
Hence it will take 15 years to reach more than
£300.
Problem 20. A drilling machine is to have 6
speeds ranging from 50 rev/min to 750 rev/ min. If
the speeds form a geometric progression determine
their values, each correct to the nearest whole
number.
Let the GP of n terms be given by a, ar, ar 2 , ..., ar n−1 .
The first term a = 50 rev/min
The 6th term is given by ar 6−1 , which is 750 rev/min,
i.e.
ar
5
= 750
from which r
5
=
750
a
=
750
50
= 15
Thus the common ratio, r =
5
√
15 = 1.7188
The first term is a = 50 rev/min
the second term is ar = (50) (1.7188) = 85.94,
the third term is ar 2 = (50) (1.7188) 2 = 147.71,
the fourth term is ar 3 = (50) (1.7188) 3 = 253.89,
the fifth term is ar
4
= (50) (1.7188)
4
= 436.39,
the sixth term is ar 5 = (50) (1.7188) 5 = 750.06
Hence, correct to the nearest whole number, the 6 speeds
of the drilling machine are 50, 86, 148, 254, 436 and
750 rev/min.
Now try the following exercise
Exercise 27 Further problems on geometric
progressions
1. In a geometric progression the 5th term is
9 times the 3rd term and the sum of the 6th
and 7th terms is 1944. Determine (a) the common ratio, (b) the first term and (c) the sum
of the 4th to 10th terms inclusive.
[(a) 3 (b) 2 (c) 59022]
2. Which term of the series 3, 9, 27, ... is
59049?
[10 th ]
3. The value of a lathe originally valued at
£3000 depreciates 15% per annum. Calculate
its value after 4 years. The machine is sold
when its value is less than £550. After how
many years is the lathe sold?
[£1566, 11 years]
4. If the population of Great Britain is 55 million
and is decreasing at 2.4% per annum, what
will be the population in 5 years time?
[48.71 M]
5. 100 g of a radioactive substance disintegrates
at a rate of 3% per annum. How much of the
substance is left after 11 years?
[71.53 g]
6. If £250 is invested at compound interest of
6% per annum determine (a) the value after
15 years, (b) the time, correct to the nearest
year, it takes to reach £750.
[(a) £599.14 (b) 19 years]
7. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the
speeds form a geometric progression determine their values, each correct to the nearest
whole number.
[100, 139, 193, 268, 373, 518,
720, 1000 rev/min]