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1. ADDITIONAL MATHEMATICS
NAME: David Ramdeen
REGISTRATION #:
SCHOOL: Wolmer’s Boys’ School
SCHOOL CENTRE#:100128
TITLE: Use of A Series/Progression in Finding Total Repayment
STUDY: Sequence and Series
PIECE: EXTERNAL SCHOOL BASED ASSESSMENT
TEACHER: Mr. Leamy

2. Table of Contents
Problem Statement……………………………….1
Mathematical Formulation……………….……...2
Solution of Problem………………………………3
Application of Solution…………………………...8
Discussion of Findings…………………..………10
Conclusion and Suggestions…………………….11

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Problem
A customer at a bank borrows a loan in order to finance a business. The customer begins to repay
the loan monthly by increasing the payment by $X. “X” is used because she had not taken note
of the increase monthly. She pays $50 in the 5th month and $70 in the 9th month. However, she
and her friend who is a loan processor (displayed below) are trying to determine the total
repayment after 24 months. Due to other taxes and expenses she wants to know how much to
save for the repayment to avoid financial pressure.
To begin they have to take two main things to take into consideration:
 The common difference pay between the monthly intervals.
 The first payment.
A. To show that they should use an arithmetic progression in order to calculate the total
repayment.
B. Depending on the progression used what was the common difference(d) and the partial
sum(Sn) ie. the total repayment

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Mathematical Formulation
A.
Formulation of Elements of the Problem Statement
 Monthly Increase (Common Difference) = $X
 Given Payments: $50 in 5th month and $70 in 9th month
 Number of Terms: 24
1. With the help of the diagram below, define an arithmetic progression
Figure.1
2. Using the diagram below differentiate between an arithmetic and a geometric
progression, additionally state why a geometric progression couldn’t be another
alternative to finding the total repayment

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Figure 2.
3. Prove that the loan processor should use formulae of arithmetic progression for the
given problem.
4. Prove that the partial sum is equal to total repayment.
5. Prove that the total repayment is = n/2[2a + (n-1) d]
Key: n- number of terms
a-first term
d-common difference
Solution of Problem
A.
To show that they should use the arithmetic progression related formulas in order to calculate the
total repayment.

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1. With the help of the diagram below, define an arithmetic progression.
Figure.3
 Assumption-We can assume that since the bar are constantly increasing by 2, an
arithmetic progression is a sequence of numbers with a common increase.
PROOF
In relation to the bar chart, each successive bar is constantly added by two on the y axis
represented by f(x). In this progression the difference would be 2. Therefore, an arithmetic
progression (AP) or arithmetic sequence is a sequence of numbers such that the difference
between the consecutive terms is constant.
2. Using the diagram below differentiate between an arithmetic and a geometric
progression, additionally state why a geometric progression couldn’t be another
alternative to finding the total repayment

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Figure 4.
 Assumption-According to the G.P. graph, it is a sequence of numbers with a
common ratio, on the other hand according to the A.P graph, it’s a sequence of
numbers with a common difference. Hence, because they’re different, if the total
repayment was calculated with both formulae, you would get different answers.
PROOF
In relation to the graph above, as shown the blue line represents a G. P (Geometric Progression)
while the red represents an A. P (Arithmetic Progression). As shown the y axis of G.P increases
rapidly, this illustrates that a G.P has no set common difference but a common ratio i.e. the
amount between each number in a geometric sequence. It is called the common ratio because it’s
a ratio between two consecutive numbers in the sequence. The common ratio of the G.P graph is
3, (2:6) or (6:18) hence each following value is multiplied by 3, comparatively, in the A.P graph
the common difference is 2 because each following value is added by 2, preceding value is
calculated by taking away 2 from the current value hence bringing about the “difference”. Due to
this difference of A.P and G.P, it’s obvious they couldn’t be substituted for another. G.P couldn’t

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be used to calculate the total repayment because it’s the sum of terms with common differences
and oppositely, G.P calculates the sum of terms with common ratios
3. Prove that the loan processor should use formulae of arithmetic progression for the
given problem.
 Assumption- The definition of A.P was stated earlier, and the problem is directly
proportional to an arithmetic progression
PROOF
In the problem, it is stated that the lady repays the loan monthly with the price increasing by an
unknown amount represented as “$X”. Therefore, “X” is the common difference which is the
constant amount by which consecutive numbers are increasing.
4. Prove that the partial sum is equal to total repayment.
 Assumption-Since partial sum is the sum of a set number of terms it is in
accordance with total repayment.
PROOF
The partial sum is the sum of a limited (that is to say, a finite) number of terms for e.g. sum of
the first 20 terms. The total repayment desired is the sum of monthly loans repaid in a sequential
order in this case the total in 24 months. Therefore, there will be 24 finite terms, after the first
month (first term e.g. $20), there is an increased value since interest would more than likely be
on the loan. After these 24 months, the total repayment is to be calculated, since, the partial sum

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(Sn) is the sum of first “n” months, we can therefore substitute these 24 months in place of n.
With help of other calculations e.g. common difference, the partial sum can be determined hence
giving the total repayment
5. Prove that the total repayment is = n/2[2a + (n-1) d]
Key: n- number of terms
a-first term
d-common difference
 Assumption-An A.P is infinite hence this formula comes about in order to find the
nth term by using the formula for finding terms
PROOF
Firstly, the formula for the“nth” term must be established.” Nth” can be 9th,4th, etc.
The “nth” term is represented by: Tn “n” being a certain number
So, a = 1st term, a + d = 2nd term, a + 2d = 3rd term,.........
Therefore Tn= a + (n-1) d
Since partial sum (Sn) is the sum of the first nth terms it can be derived from the method
below(eq.1):
Sn= a + (a + d) + (a + 2d) ...........+ a+(n-1)d
If it was represented backwards it would be(eq.2):
Sn= a+(n-1) d + a+(n-2) d………+ a
↑- (n-2) is used because it is backwards e.g. 100,99,98
Then eq.1 and eq.2 are added.
2Sn=[a+a+(n-1) d] + [(a+d)+a+(n-2) d] +....................

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2Sn=[2a+(n-1) d] + [2a+(n-1) d] +........
Therefore, in total there are “n” amount of [2a+(n-1) d]
2Sn= n[2a+(n-1) d]
Therefore Sn= n/2[2a + (n-1) d]
Application of Solution
B.
Depending on the progression used what was the common difference(d) and the partial sum(Sn)
ie. the total repayment
Common Difference
1st Pay= a
2nd pay=a + x (monthly increase)
Therefore,5th pay = a+4x which was $50 and the 9th pay=a+8x which was $70
The method of simultaneous equations can now be used to acquire the common difference and
hence the first term.
a+4x=50
a+8x=70
If we subtract we would get: -4x= -20 therefore “x” ie. the common difference = 5 hence we can
determine the first term by means of substituting x into any of the above equations:
a+4(5) =50
a+20=50 Hence, “a” ie. the first term =30

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As a result, the total repayment can now be solved by means of the partial sum/summation
formula.
Sn=n[2a+(n-1) d]
Sn=24/2[2(30) +(24-1)5]
Sn=12[60+(23)5]
Sn=12[60+115]
Sn=12[175]
Sn=2100
Therefore, the total repayment after 24 months was $2100.

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Discussion of Findings
Based on previous data analysis and calculations we have shown that an arithmetic progression was the
accurate sequence to deduce the monthly loan increase and thus total repayment ,hence it was proven that
the loan processor should use the formula below to derive total repayment = n/2[2a+(n-1)d]so he can aid
his friend in avoiding financial pressure .Consequently, in order to calculate the total repayment we
calculated the common difference(d) to be $5,and the first term/first payment(a) to be $30.With that being
the case the total repayment after 24 months was $2100.

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Conclusion and Suggestions
After all the data processing, it can be inferred that namely, an arithmetic progression and other
Additional Mathematics related topics has many uses to the real world. As a matter of fact, that
could be a reason for this SBA, to prove that these topics have uses, because students are
constantly questioning the purpose of the subject. In other respects, I must articulate that this
School Based Assessment has had another benefit which was improving my problem-solving and
critical thinking.
Note that if finance and commerce is a commonly related topic in your life, it is recommended
that a geometric progression is used to pay mortgages, loans, investments, etc. as these are
money-centered tasks