Higher Engineering Mathematics, Sixth Edition

54 Higher Engineering Mathematics

Problem 10. An oil company bores a hole 80m

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=2andn= 80

Thus,total cost,

Sn=

n

2

[

2 a+(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 progres-
   sion 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
   progressionare9,40.5,and425.5respectively.
   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 £16000per 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 12years.
   [£19840, £223,680]
   8. An oil company bores a hole 120m deep. Esti-
   mate the cost of boring if the cost is £70 for
   drilling the first metre with an increase in cost
   of £3per metre for each succeeding metre.
   [£29820]

6.4 Geometric progressions

When a sequence has a constant ratio between succes-

sive terms it is called ageometric progression(often

abbreviated to GP). The constant is called thecommon

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 isr.

General expression for then’th term of a GP

If the first term of a GP is ‘a’ and the common ratio is

r,then

then’th term is:arn−^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 ,sincea=1andr=2.

Sum ofnterms of a GP

Let a GP bea,ar,ar^2 ,ar^3 ,...,arn−^1

then the sum ofnterms,

Sn=a+ar+ar^2 +ar^3 +···+arn−^1 ··· (1)

Multiplying throughout byrgives:

rSn=ar+ar^2 +ar^3 +ar^4

+···+arn−^1 +arn+··· (2)

Subtracting equation (2) from equation (1) gives:

Sn−rSn=a−arn

i.e. Sn( 1 −r)=a( 1 −rn)

Thus the sum ofnterms,Sn=a(1−r

n)

(1−r) which is valid

whenr<1.