basic_engineering_mathematics0

222 Basic Engineering Mathematics
Now try the following exercise
Exercise 106
Further problems on arithmetic
progressions (Answers on page 281)
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.
2. Three numbers are in arithmetic progression. Their sum
is 9 and their product is 20 1 . Determine the three
4
numbers.
3. Find the sum of all the numbers between 5 and 250 which
are exactly divisible by 4.
4. Find the number of terms of the series 5, 8, 11, ... of
which the sum is 1025.
5. Insert four terms between 5 and 22 1 to form an arithmetic
progression.
2
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.
7. On commencing employment a man is paid a salary
of £7200 per annum and receives annual increments of
£350. Determine his salary in the 9th year and calculate
the total he will have received in the first 12 years.
8. 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.
29.6 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
If the first term of a GP is ‘a’ and the common ratio is r, then
the n ′ th term is : 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:
rS n = ar + ar 2 + ar 3 + ar 4 +···ar n−1 + ar n ··· (2)
Subtracting equation (2) from equation (1) gives:
S n − rS n = a − ar n
i.e. S n (1 − r) = a(1 − r n )
Thus the sum of n terms, S n = a(1 − rn )
(1 − r)
when r < 1
Subtracting equation (1) from equation (2) gives
S n = a(rn − 1)
(r − 1)
which is valid when r > 1
which is valid
For example, the sum of the first 8 terms of the GP 1, 2, 4,
8, 16, ...is given by
S 8 = 1(28 − 1)
, since a = 1 and r = 2
(2 − 1)
1(256 − 1)
i.e. S 8 = = 255
1
When the common ratio r of a GP is less than unity, the sum of
n terms,
S n = a(1 − rn )
, which may be written as
(1 − r)
S n =
a
(1 − r) − arn
(1 − r)
Since r < 1, r n becomes less as n increases,
i.e. r n → 0 as n →∞
ar n
Hence → 0 as n →∞
(1 − r)
Thus S n → a as n →∞
(1 − r)
a
The quantity
(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 < r < 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
Let a GP be a, ar, ar 2 , ar 3 , ...ar n−1
For example, the sum to infinity of the GP 1, 1 2 , 1 4 , ...is
S ∞ = 1
1 − 1 , since a = 1 and r = 1 2 , i.e. S ∞ = 2
2