Appendix For Online Publication
A.1 Dataset description
3.2 Series
3.3 Arithmetic Progression (AP) 3.4 Geometric Progression (GP) 4.0 Conclusion
5.0 Summary
6.0 Tutor-Marked Assignment 7.0 References/ Further Readings
1.0 INTRODUCTION
Sequence of numbers — when a set of numbers is arranged in a recognizable pattern so that there is a first, second and third number and so on, then the set of numbers is called a sequence of numbers, e.g. 2, 4, 6, 8, ... Such numbers are arranged according to a definite rule and each number in the sequence is called a term of the sequence. The terms are usually expressed as xl, x2, x3... x„ ... where x, is the first term, x2 is the second term, etc. The nth term x„, is the general term. The two kinds of sequences useful for business applications, especially in financial and banking (interest) calculations are: arithmetic progression (arithmetic sequence) and geometric progression (geometric sequence).
2.0 OBJECTIVES
After studying this unit, you should be able to:
define the arithmetic and geometric progression;
calculate sequences of numbers in arithmetic or geometric progression in business;
3.0 MAIN CONTENT 3.1 Sequences
If for every positive integer (number) n, there corresponds a number an is related to n by some rule, then the terms a
l, a
2...,
an ...are said to form a sequence. A sequence is denoted by bracketing its nth term i.e.
(a
n).
MBE 839 QUANTITATIVE TECHNIQUES FOR BANKING AND FINANCE
Example of some sequences:
(i) If an = n2, then sequence (an) is 1, 4, 9, 16, .... n2, ....
(ii) If an = 1/n, then sequence (a„) is I, 1/4, 1/4, 1/4, ...., 1/n ...
(iii) If an = n2/n+1, then sequence (an) is 1/4, 4/3, 9/4, ...., n2/n+1, Note: The concept of sequence is very useful in finance and banking —
simple and compound interest, annuities, present value and mortgage payments.
3.2 Ser ie s
A series is formed by connecting the terms of a sequences connoting plus or minus sign. Thus, if an is the nth term of a sequence, then: a, + a2 ... an is a series of n terms.
3.3 Ar it hmet ic Pro gress io n ( AP)
A progression is a sequence whose successive terms indicate the growth or progress of some characteristics. An arithmetic progression is a sequence whose term increases or decreases by a constant number — common difference of an A.P, denoted by d. That is, each term of the arithmetic progression after the first is obtained by adding a constant d to the preceding term. The standard form of an A.P is written as:
a, a + d, a + 2d, + a + 3d.
where 'a' is called the first term. Thus, the corresponding standard form of an arithmetic series becomes:
a + (a + d) + (a + 2d) + (a + 3d) + Example I
Suppose we invest N100 at a simple interest of 15% per annum for 5 years. The amount at the end of each year is given by 115, 130, 145, 160, 175.
This forms an arithmetic progression.
The nth term:
The nth term of an A.P is also called the general term of the standard A.P. It is given by:
MBF 839 QUANTITATIVE ma INIQUES FOR BANKING AND FINANCE
Consider the first n terms of a, a + d, a + 2d, a + 3d ..., a + (n — 1) d.
The sum, S,, of these terms is given by:
= a + (a + d) + (a + 2d) + (a + 3d + + a + (n — 1) d
= (a -I a ) ... 1- a) F d {1 + 2 -I- 3 I ....± (n — 1){
= n.a + d n (n — 1) } (using formula for the sum of first (n— 1) 2 natural numbers)
n { 2a (n — 1) cl}
2 Example 2
Suppose Mr. Sola repays a loan of N3,250 by paying N20 in the first month and then increases the payment by N15 every month. How long will be taken to clear his loan?
Solution:
Since Mr. Sola increases the monthly payment by a constant amount, N15 every month, therefore d = 15 and first month installment is, a = N20. This forms an A.P. If the entire amount be paid in a monthly installment, then we have:
n{ 2a+(n— 1)d}
2
or 3,250 = n { 2 x 20 + (n — 1) 15}
2
3,250 x 2 = n { 2 x 20 + (n — 1) 15}
6,500 = n{25 + 15n }
15n2 -I- 25n —6500 = 0 a quadratic equation in n.
Find n applying formula as previously discussed in unit 2.
—b ± b2 — 4ac
2a
—25 ± 4(25)2— 4 x 15 x ( — 6500) 2x 15
—25 ± 625 = 20 or —21.66 30
MBE 839 QUANTITATIVE TECHNIQUES FOR BANKING AND FINANCE
The value n = —21.66 (negative, hence meaningless). Mr Sola will pay the entire amount in 20 months.
SELF ASSESSMENT EXERCISE 2
(i) Find the 15th term of an A.P. whose first term is 12 and common difference is 2.
(ii) A retailer's revenue and costs exactly balanced on December 31, 2001. During the following year, his costs rose by N150 each month and his revenue rose by N200 each month. What were; (a) the cost incurred and (b) the revenue received during the month of December 2002?
3.4 Geometric Progression (G.P)
A geometric progression (G.P) is a sequence when each terms increases or decreases by a constant ratio called common ratio of G.P and is denoted by r. In other words, each term of G.P is from the first by multiplying the preceding term by a constant r. The standard form of a G.P is expressed as a, ar, ar2..., where `a' is called the first term.
Thus, the corresponding geometric series in standard form is a + ar + ar2 +
Example 3
Suppose we invest 14100 at a compound interest of 12% per annum for three years. The amount at the end of each is calculated as follows:
(i) Interest at the end of first year:
100 x 12 =- 1412
100
Amount at the end of first year:
Principal + Interest
100 + 100 (12/100)
100 0 + 12/100)
33
MBE 839 QUANTITATIVE TECI INIQUFS FOR BANKING AND FINANCE
(Principal at the beginning of second year) = { 1 + 1 2 }
100 (1 + 12/100) {1 + 12/100) 100
100 {1 + 12/100}2
Amount at the end of third year:
100 {1 + 12/100) {1 + 12/100) {1 + 12/100}
100 {1 + 12/100}3
Thus, the progression giving the amount at the end of each year is:
100 {1 ±12/100); 100 (1 ±12/100)2; 100 {1 +12/1O0};
This is a G.P with common ratio r = {1 -I- 12/100)
In general, if P is the principal and the compound interest rate per annum, then the amount at the end of first year becomes P (1 + 1/100).
Also, the amount at the end successive years forms a G.P.
= P(1 + 1/100); P (1 + 1/100)2; ... with r = P (1 -I- 1/100)
The nth term of:
The nth term of G.P. is called the general term of the standard G.P. It is given by
Tn arn — I, n = 1, 2, 3....
Note here that the power of r is one less than the index of Tn, which denotes the rank of this term in the progression.
Sum of the First n Terms in G.P
Consider the first n terms of the standard form of G.P.
a, ar, ar2, arn- '
MB]: 839 QUANTITATIVE TECIINIQUES FOR BANKING AND FINANCE
The sum, S„ of these terms is given by:
s„
a + ar +ar
2 arn 2 + a?' I ... (1) Multiplying both sides by r, we get:rS„ = ar -F ar2 ar3 + + arn-2 -I- a? (2) Subtracting (2) from (1), we have:
S„, — rS„ a —
S„ (1 —r) a (1 —?) Or S = a (1 — ; r #1 and < 1
(1 — r)
Changing the signs of the numerator and denominator, we have:
S„= a (rn — 1) ; r #1 and > 1 (r — 1)
(a) If r = 1, G.P becomes a, a, a, ... so that S„ in this case is Sr, = n.a.
(b) If number of terms in a G.P are infinite, then:
S„= a ; r < 1
(1 — r) For r I, the sum tends to infinity.
Example 4
A car is purchased for N80,000. Depreciation is calculated at 5% per annum for the first 3 years and 10% per annum for the next 3 years.
Find the money value of the car after a period of 6 years.
Solution:
(i) Depreciation for the first year:
80,000 x 5 100
The depreciated value of the car at the end of first year is:
80,000 — 80,000 x 5 100 80,000 (1 — 5 )
MBE 839 QUANTITATIVE TED INIQUES FOR BANKING AND FINANCE
( ii) D e pr e c ia t ion f or t he s e c ond ye a r :
(Depreciated value at the end of first year) x (rate of depreciation for second year)
80,000(1 — 5 ) 5 100 100
Thus, the depreciated value at the end of the second year is:
(depreciated value after first year) — (depreciation for second year)
80,000 80,000 80,000
(1 0 (1
— — 80,000 100
— 5) 0 — 5) 100 100
— 5)2 100
(1 — j) 100 /51
100
Calculating in the same way, the depreciated value at the end of three years is:
(depreciated value after second year) — (depreciation for third year)
80,000(1 — 5) — 80,000 (1 —5) (5) (5)
100 100 100 100
80,000 (1 — 5)0 —5) (1
-5)
100 100 100 80,000(1 _)3
100
( i i i ) D e p r e c ia t i o n f or f ou r t h ye a r : 80,000(1 _)3 (10)
100 100
Thus, the depreciated value at the end fourth year is:
(depreciation value after three years) x (depreciation for fourth year)
MBE 839 QUANTITATIVE TECHNIQUES FOR BANKING AND FINANCE
80,000(1 _)3 — 80,000 (1 _)3 (L))
100 100 100
Calculating in the same way, the depreciated value at the end of six years becomes:
= 80,000(1 —5f — 80,000 (1 _)3 (1 0)3
100 100 10
N49,980.24
SELF ASSESSMENT EXERCISE 3
(i) Find the 6th term of the geometric progression 72, —24, 8, ....
(ii) You decide to increase your saving by N200 per annum. If you save N500 in the first year, how long will it take to save N5,300?
4.0 CONCLUSION
In this unit, you can define sequences of numbers in arithmetic or geometric progression and examples of their application in some business calculations.
5.0 SUMMARY
This unit has directed you to define, distinguish and establish managerial applicability of arithmetical and geometrical progressions in business operations.
6.0 TUTOR
-MARKED ASSIGNMENTS
(1) Find the general term and the sum of first 20 terms, of the arithmetic progression 2, 6, 10 ...
(2) Suppose that people spent 0.9 and save 0.1 of any money received. If N1,000 is received, then N900 is spent and N100 saved. The people receiving the N900 will spend N810 and save N90. This process will continue. Find the maximum sum of a geometric progression.
MBE 839 QUANTITATIVE TECIINIQUES FOR BANKING AND FINANCE
7.0 REFERENCES AND FURTHER READINGS
Indira Gandhi National Open University (2000). School of Management Studies. MS — 8 Quantitative Analysis for Managerial Applications, Basic Mathematics for Management.
Owen, F. and Jones, R. (1975). Modern Analytical Techniques, Polytech Publishers Limited, Stockport.
National Open University of Nigeria (2004). MBA 717: Basic Mathematics and Statistics.
MBF 839 QUANTITATIVE TECHNIQUES FOR BANKING AND FINANCE
UNIT 4 INVESTMENT APPRAISAL I