54 Higher Engineering Mathematics
Problem 10. An oil company bores a hole 80 m
deep. Estimate the cost of boring if the cost is £30
for drilling the first metre with an increase in cost of
£2 per metre for each succeeding metre.
The series is: 30, 32, 34, ... to 80 terms, i.e. a = 30,
d = 2 and n = 80
Thus, total cost,
S n =
n
2
2a + (n − 1)d
=
80
2
[2(30) + (80 − 1)(2)]
= 40[60 + 158] = 40(218) = £8720
Now try the following exercise
Exercise 25 Further problems on arithmetic
progressions
1. The sum of 15 terms of an arithmetic progression is 202.5 and the common difference is 2.
Find the first term of the series.
[−0.5]
2. Three numbers are in arithmetic progression.
Their sum is 9 and their product is 20.25.
Determine the three numbers.
[1.5, 3, 4.5]
3. Find the sum of all the numbers between 5 and
250 which are exactly divisible by 4. [7808]
4. Find the number of terms of the series 5, 8,
11, ... of which the sum is 1025.
[25]
5. Insert four terms between 5 and 22.5 to form
an arithmetic progression. [8.5, 12, 15.5, 19]
6. The first, tenth and last terms of an arithmetic
progression are 9, 40.5, and 425.5 respectively.
Find (a) the number of terms, (b) the sum of
all the terms and (c) the 70th term.
[(a) 120 (b) 26070 (c) 250.5]
7. On commencing employment a man is paid
a salary of £16000 per annum and receives
annual increments of £480. Determine his
salary in the 9th year and calculate the total
he will have received in the first 12 years.
[£19840, £223,680]
8. An oil company bores a hole 120 m deep. Estimate the cost of boring if the cost is £70 for
drilling the first metre with an increase in cost
of £3 per metre for each succeeding metre.
[£29820]
6.4 Geometric progressions
When a sequence has a constant ratio between successive terms it is called a geometric progression (often
abbreviated to GP). The constant is called the common
ratio, r.
Examples include
(i) 1, 2, 4, 8, ... where the common ratio is 2 and
(ii) a, ar, ar 2 , ar 3 , ... where the common ratio is r.
General expression for the n’th term of a GP
If the first term of a GP is ‘a’ and the common ratio is
r, then
the n’th term is: ar
n−1
which can be readily checked from the above examples.
For example, the 8th term of the GP 1, 2, 4, 8, ... is
(1)(2) 7 = 128, since a = 1 and r = 2.
Sum of n terms of a GP
Let a GP be a, ar, ar 2 , ar 3 , ..., ar n−1
then the sum of n terms,
S n = a + ar + ar
2
+ ar
3
+ · · · + ar
n−1
· · · (1)
Multiplying throughout by r gives:
r S n = ar + ar
2
+ ar
3
+ ar
4
+ · · · + ar
n−1
+ ar
n
+ · · ·
(2)
Subtracting equation (2) from equation (1) gives:
S n − r S n = a − ar
n
i.e. S n (1 − r) = a(1 − r
n
)
Thus the sum of n terms, S n =
a(1 −r n )
(1 − r ) which is valid
when r < 1.
Problem 10. An oil company bores a hole 80 m
deep. Estimate the cost of boring if the cost is £30
for drilling the first metre with an increase in cost of
£2 per metre for each succeeding metre.
The series is: 30, 32, 34, ... to 80 terms, i.e. a = 30,
d = 2 and n = 80
Thus, total cost,
S n =
n
2
2a + (n − 1)d
=
80
2
[2(30) + (80 − 1)(2)]
= 40[60 + 158] = 40(218) = £8720
Now try the following exercise
Exercise 25 Further problems on arithmetic
progressions
1. The sum of 15 terms of an arithmetic progression is 202.5 and the common difference is 2.
Find the first term of the series.
[−0.5]
2. Three numbers are in arithmetic progression.
Their sum is 9 and their product is 20.25.
Determine the three numbers.
[1.5, 3, 4.5]
3. Find the sum of all the numbers between 5 and
250 which are exactly divisible by 4. [7808]
4. Find the number of terms of the series 5, 8,
11, ... of which the sum is 1025.
[25]
5. Insert four terms between 5 and 22.5 to form
an arithmetic progression. [8.5, 12, 15.5, 19]
6. The first, tenth and last terms of an arithmetic
progression are 9, 40.5, and 425.5 respectively.
Find (a) the number of terms, (b) the sum of
all the terms and (c) the 70th term.
[(a) 120 (b) 26070 (c) 250.5]
7. On commencing employment a man is paid
a salary of £16000 per annum and receives
annual increments of £480. Determine his
salary in the 9th year and calculate the total
he will have received in the first 12 years.
[£19840, £223,680]
8. An oil company bores a hole 120 m deep. Estimate the cost of boring if the cost is £70 for
drilling the first metre with an increase in cost
of £3 per metre for each succeeding metre.
[£29820]
6.4 Geometric progressions
When a sequence has a constant ratio between successive terms it is called a geometric progression (often
abbreviated to GP). The constant is called the common
ratio, r.
Examples include
(i) 1, 2, 4, 8, ... where the common ratio is 2 and
(ii) a, ar, ar 2 , ar 3 , ... where the common ratio is r.
General expression for the n’th term of a GP
If the first term of a GP is ‘a’ and the common ratio is
r, then
the n’th term is: ar
n−1
which can be readily checked from the above examples.
For example, the 8th term of the GP 1, 2, 4, 8, ... is
(1)(2) 7 = 128, since a = 1 and r = 2.
Sum of n terms of a GP
Let a GP be a, ar, ar 2 , ar 3 , ..., ar n−1
then the sum of n terms,
S n = a + ar + ar
2
+ ar
3
+ · · · + ar
n−1
· · · (1)
Multiplying throughout by r gives:
r S n = ar + ar
2
+ ar
3
+ ar
4
+ · · · + ar
n−1
+ ar
n
+ · · ·
(2)
Subtracting equation (2) from equation (1) gives:
S n − r S n = a − ar
n
i.e. S n (1 − r) = a(1 − r
n
)
Thus the sum of n terms, S n =
a(1 −r n )
(1 − r ) which is valid
when r < 1.