MATHEMATIC 1 4 5 Klasifikasi Engine
Introduction
Title:Interest rate and credit card debt
Interest is payment from a borrower which more than
the money we borrow. Interest is the charge for the
privilege of borrowing money, typically expressed as
annual percentage rate. Interest can also refer to the
amount of ownership a stockholder has in a
company, usually expressed as a percentage like
1.5%,or a decimal 0.015.
There are two main types of interest that can be
applied to loans: simple and compound. Simple
interest is a set rate on the principle originally lent to
the borrower that the borrower has to pay for the
ability to use the money. Compound interest is
interest on both the principle and the compounding
interest paid on that loan. The latter of the two types
of interest is the most common.
Interest rates have a huge efect on loans. In short,
loans with high interest rates have higher monthly
payments or take longer to pay of than loans with
low interest rates.
In mathematics, a geometric series, also known as a
geometric sequence, is a sequence of numbers
where each term after the frst is found by
multiplying the previous one by a fxed, non-zero
number called the common ratio.
Each credit card has a maximum limit. A maximum
limit of RM20000 means total money spent plus
interests incurred must not exceed this amount. This
can ensure that the credit card users will not spent
too much money by using credit card. Each credit
card user will choose a payment plan to repay the
debt every month.
Methodology
i=
Annual interest rate
12
Annual interest rate =18%
i = monthly interest rate
We use 18% divided 12 and get a answer 1.5% .1.5% is our monthly
interest rate.
Month,
n
Interest
due
(RM)
0
1
Payment,
R
(RM)
_
120
Paid to
principal
2
520
3
520
4
520
5
520
6
520
(RM)
(RM)
_
520
Balance,bn
_
400
Interest due = Balance,bn ×1.5%
Paid to principal = Payment,R - Interest due
8000
7600
Balance,bn = Balance,bn-1 - Paid to principal
bn =arn-1
a =8000
Write equation for b1,b2,b3,fnd the frequency and relationship
amog them by compare their diference and make a formula
for bn
For example: equation b1 is b1 =b0 – [ R-(b0)(i)]
Find the equation forb2 and b3 with equation b1.
make three table with three amont larger than 520 such as
600.800.1000 to fnd monthly balance until balance bn is
less than payment.The last month,n will be the month paid
of the payment
Interest due = Balance,bn ×1.5%
Paid to principal = Payment,R - Interest due
Balance,bn = Balance,bn-1 - Paid to principal
bn =arn-1
a =8000
Payment= 520,600,800,1000
When bn =0,
0= bn-1 – [ R-(bn-1)(i)]
b1 =b0 – [ R-(b0)(i)]
We can sub n=0 to fnd equation about b0
Use equation b1 =b0(1+i),b2,b3 and bn.Show bn = b0(1+i)
s
(1+i+ 100 )n-1
b2=b0(1+i)(1+i+ s )
100
b3=b0(1+i)(1+i+ s )2
100
Try to fnd relationship among b1, b2,b3 ,to form a equation for
bn
1
When s=1,sub s=1 into equation bn = b0(1+i)(1+i+ 100 )n-1
1
bn = b0(1+i)(1+i+ 100 )n-1
Then ,sub bn =16000
b0 =8000
i= 0.015
1
16000=8000(1+0.015)(1+0.015 + 100 )n-1
Solve the equation to fnd n
Result
monthly interest rate
i=
Annual interest rate
12
i = 18%
12
i = 1.5%
i=monthly interest rate =1.5%
Every mont ,we will be charge 1.5% of interest rate from our
balance.
This interest rate make us paid more mony then
beginning.Interest rate is also a way to let bank earn money.
When payment =RM520
Month, Interest Payment, Paid to Balance,
n
due
R
princip bn
al
(RM)
(RM)
(RM)
(RM)
0
1
2
3
4
5
6
7
_
120
114
107.91
101.73
95.45
89.09
_
520
520
520
520
520
520
520
82.62
8
9
10
520
76.06
69.4
520
520
62.64
11
520
55.78
12
13
48.
82
520
520
_
400
406
412.09
418.27
424.55
430.91
437.3
8
443.9
4
450.6
457.3
6
464.2
2
8000
7600
7194
6781.9
6363.6
5939.1
5508.2
5070.82
4626.88
4176.28
3718.92
3254.7
471.1
8 2783.52
478.2
5 2305.27
485.4
2 1819.85
492.7 1327.15
500.0
9 827.06
507.5
9 319.47
41.75
14
15
16
520
34.58
27.30
520
520
19.91
17
520
12.41
When payment =RM520,month balance paid of =17
At 17 th month,payment from RM 520 increase to RM
839.47.This make the payment faster paid end and save the
interest rate charge by the bank. At 17th month , balance is
fully paid of.
When payment =RM 600
Month,
n
Interest Payment,R Paid to
Balance,bn
due
principal
(RM)
0 _
_
1
120
2
112.8
3 105.49
4
98.07
5
90.55
6
82.9
7 75.15
8 67.28
9 59.29
10 51.18
11 42.94
12 34.59
(RM)
(RM)
_
600
600
600
600
600
600
600
600
600
600
600
600
480
487.2
494.51
501.93
509.45
517.1
524.85
532.72
540.71
548.82
557.06
565.41
(RM)
8000
7520
7032.8
6538.3
6036.4
5526.9
5009.8
4485.15
3952.43
3411.72
2862.9
2305.84
1740.43
13
14
26.11
17.50
600
600
573.89
582.5
1166.54
584.04
When payment =RM600,month balance paid of =14
At 14 th month,payment from RM 600 increase to RM
1184.04This make the payment faster paid end and save the
interest rate charge by the bank. At 14th month , balance is
fully paid of.
When payment =RM 650
Month,
n
Interest Payment,R Paid to
Balance,bn
due
principal
(RM)
(RM)
0 _
_
1
120
2 112.05
3 103.98
4 95.79
650
650
650
650
5
87.48
650
6
79.04
650
_
650
7
8
70.48
650
61.78
(RM)
530
537.95
546.02
554.2
1
562.5
2
570.9
6
579.5
2
588.2
2
(RM)
8000
7470
6932.05
6386.03
5831.8
2
5269.3
0
4698.3
4
4118.8
2
3530.6
0
9
650
52.96
10
650
44
11
650
34.91
12
650
25.69
13
650
16.32
597.0
4
2933.5
6
2327.5
6
1712.4
7
1088.1
6
606
615.0
9
624.3
1
633.6
8 454.48
When payment =RM650,month balance paid of =13
At 13 th month,payment from RM 650 increase to RM
1094.48 .This make the payment faster paid end and save
the interest rate charge by the bank.At 13 th month , balance
is fully paid of.
When payment =RM800
Month,n Interes
t due
(RM)
Payment,R Paid to Balance,bn
principal
(RM)
(RM)
(RM)
0 _
_
1
120
2
109.8
3
99.45
4
88.94
5
78.27
6
67.45
7 56.46
8 45.31
9 33.99
10
22.5
_
800
800
800
800
800
800
800
800
800
800
680
690.20
700.60
711.10
721.70
732.60
743.50
754.70
766.01
777.50
8000
7320
6629.80
5929.30
5218.20
4496.50
3763.90
3020.50
2265.80
1499.80
722.30
When payment =RM800,month balance paid of =10
At 10th month,payment from RM 800 increase to RM1522.30
This make the payment faster paid end and save the interest
rate charge by the bank. At 10th month , balance is fully paid
of.
When payment =RM1000
Month,n Interest Payment,R Paid to Balance,bn
due
principal
(RM)
(RM)
(RM)
(RM)
0 _
_
_
8000
1
120
1000
880
7120
2
106.8
1000
893.2
6226.8
3
93.4
1000
906.6
5320.2
4
79.8
1000
920.2
4400
5
66
1000
934
3466
6
51.99
1000 948.01
2517.99
7 37.77
1000
51.99
2466
8 36.99
1000 963.01 1502.99
9 22.54
1000 977.46
525.53
When payment =RM1000,month balance paid of =9
At 9 th month,payment from RM 1000 increase to
RM1525.53.This make the payment faster paid end and save
the interest rate charge by the bank. At 9 th month , balance
is fully paid of.
b1 =b0 – [ R-(b0)(i)] b1 = b0 – ( R – ib0 )
= ( 1 + i )b0 – R
b2 = b1 – ( R – ib1 )
= ( 1 + i )b1 – R
Substitute b1 into b2 , b2 = ( 1 + i ) [ ( 1 + i )b0 – R] – R
= ( 1 + i ) 2b0 - ( 1 + i )R – R
b3 = b2 – ( R – ib2 )
= ( 1 + i )b2 – R
Substitute b2 into b3 , b3 = ( 1 + i ) [ ( 1 + i )2b0 - ( 1 + i )R –
R]–R
= ( 1 + i ) 3b0 - ( 1 + i )2R - ( 1 +
i )R – R
After expressed b1 , b2 and b3 in term of R, i, n, and b0, I
observe relationship between them to form a formula for bn.
bn = b0( 1 + i )n - ( 1 + i )n-1R - ( 1 + i )n-2R - … - ( 1 + i )R – R
= b0( 1 + i )n– R [( 1 + i )n-1 + ( 1 + i )n-2 + … + ( 1 + i ) – 1]
= b0( 1 + i )n– R [1 + ( 1 + i ) + ( 1 + i )2 + ( 1 + i )3 + … +
( 1 + i )n-1]
Formula of sum:
a=1
, r=1+i
a(r n−1)
Sn = r−1
1[(1+i)−1]
= (1+i)−1
(1+i)n−1
=
i
Formula for bn
n−1]
bn = b0( 1 + i )n– R [(1+i)¿¿
¿
i
bn =balance in nth month
R =Payment
i = montly interest rate
n = month
bn = b0( 1 + i )n–
R [(1+i)¿¿ n−1]
¿
i
When payment = RM 520
n−1]
bn =8000 ( 1 + 0.015 )n– 520[(1+0.015)¿¿
¿
0.015
bn =0
8000 ( 1 + 0.015 )n–
520[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 520[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 520[( 1+ 0.015)¿¿ n−1] ¿ =0
n
120(1+0.015) – 520 ¿ ¿ =- 520
(1+0.015)n [120-520] = -520
n
(1+0.015) [(-400)] =-520
(1+0.015)n = 1.3
n log10 1.015 = log10 1.3
log 1.3
n= log 1.015
n =17.62
The month when balance is paid of while payment RM520 is
about 18 month
When payment = RM 600
bn =8000 ( 1 + 0.015 )n–
600[(1+0.015)¿¿ n−1]
¿
0.015
bn =0
n−1]
8000 ( 1 + 0.015 )n– 600[(1+0.015)¿¿
¿ =0
0.015
120(1+0.015)n – 600[(1+0.015)¿¿ n−1]
¿ =0
0.015
n
120(1+0.015) – 600 [(1+ 0.015)¿¿ n−1]¿ =0
120(1+0.015)n – 600 ¿ ¿ =-600
n
(1+0.015) [120-600] = -600
(1+0.015)n[(-480)] =-600
n
(1+0.015)
= 1.25
n log10 1.015 = log10 1.25
log 1.25
n= log 1.015
n =14.99
The month when balance is paid of while payment RM520 is
about 15 month
When payment = RM 650
n−1]
bn =8000 ( 1 + 0.015 )n– 650[(1+0.015)¿¿
¿
0.015
bn =0
8000 ( 1 + 0.015 )n–
650[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 650[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 65 0 [(1+ 0.015)¿¿ n−1]¿ =0
n
120(1+0.015) – 650 ¿ ¿ =- 650
(1+0.015)n [120-650] = -650
n
(1+0.015) [(-530)] =-650
(1+0.015)n = 1.23
n log10 1.015 = log10 1.23
log 1.23
n= log 1.015
n =13.90
The month when balance is paid of while payment RM520 is
about 14 month
When payment = RM 800
bn =8000 ( 1 + 0.015 )n–
800[(1+0.015)¿¿ n−1]
¿
0.015
bn =0
n−1]
8000 ( 1 + 0.015 )n– 800[(1+0.015)¿¿
¿ =0
0.015
120(1+0.015)n – 520[(1+0.015)¿¿ n−1]
¿ =0
0.015
n
120(1+0.015) – 800 [(1+ 0.015)¿ ¿ n−1]¿ =0
120(1+0.015)n – 800 ¿ ¿ =- 800
n
(1+0.015) [120-800] = -800
(1+0.015)n[(-680)] =-800
n
(1+0.015)
= 1.18
n log10 1.015 = log10 1.18
log 1.18
n= log 1.015
n =11.11
The month when balance is paid of while payment RM520 is
about 11 month
When payment = RM 1000
n−1]
bn =8000 ( 1 + 0.015 )n– 1000[(1+0.015)¿¿
¿
0.015
bn =0
8000 ( 1 + 0.015 )n–
1000[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 1000[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 1000[(1+ 0.015)¿¿ n−1]¿ =0
n
120(1+0.015) – 1000 ¿ ¿ =- 1000
(1+0.015)n [120-1000] = -1000
n
(1+0.015) [(-880)] =-1000
(1+0.015)n = 1.14
n log10 1.015 = log10 1.14
log 1.14
n= log 1.015
n =8.8
The month when balance is paid of while payment RM520 is
about 9 month
bn = b0( 1 + i )n-
R [(1+i)¿¿ n−1]
¿
i
If bn = 0 ,
n−1]
b0( 1 + i )n - R [(1+i)¿¿
¿ =0
i
b 0(1+i)n (i)−R [(1+i)¿¿ n−1]
¿ =0
i
b0( 1 + i )n(i) – R(1 + i)n + R = 0
b0( 1 + i )n(i) = R( 1 + i )n – R
b0 =
R [(1+i)¿¿ n−1]
¿
i(1+i)n
R
R
b0 = i - i(1+i)n
b0 =R(i)-1- R[i(1+i)n]-1
If bn = 0 , there has an expression for b0 in terms of i, n and
R.
b1=b0(1+i)
b2=b0(1+i)(1+i+ s )
100
b3=b0(1+i)(1+i+ s )2
100
b4=b0(1+i)(1+i+ s )3
100
b5=b0(1+i)(1+i+ s )4
100
b6=b0(1+i)(1+i+ s )5
100
bn=b0(1+i)(1+i+ s )n-1
100
When S=1
1
bn = b0(1+i)(1+i+ 100 )n-1
1
16000=8000(1+0.015)(1+0.015 + 100 )n-1
41
400
( 4 o )n-1 = 203
41
400
(n-1) log10 ( 4 o )= log10 203
(n-1) =27.5
n=28.4589
n=28.46
When S=2
2
bn = b0(1+i)(1+i+ 100 )n-1
2
16000=8000(1+0.015)(1+0.015 + 100 )n-1
(
207
200
400
)n-1 = 203
207
400
(n-1) log10 ( 200 )= log10 203
(n-1) =19.933
n= 20.933
When 1% is charge,Current payment is doubled after 28
month.
When 2% is charge,Current payment is doubled after 20
month.
Conclusion
In conclusion,the higher the montly
payment,the faster the time of balance paid
of,the decrease the month of the balance paid
of.If the man began and continue paid of the
current balance ,the balance can paid of
faster.If the man decide not to pay at all,the
balance will become more and have to pay
more at future.When the number of s%
increase , month when current payment double
increase.
Furthermore, interest rate will increase when
we need longer time to pay of our
payment.Balance we pay will decrease while
intrest increase through geometry series .We
need to used the shorter time we can to pay of
our payment to reduce the charge of
interest.The higher the interest rate ,the higher
the debt.The shorter the month to paid of the
debt,the less the total amount we paid .As a
consumer,we need to be wise when paying our
debt.We need to have a good planning to pay
of the debt to avoid debt become more until
we cannot repay.
Title:Interest rate and credit card debt
Interest is payment from a borrower which more than
the money we borrow. Interest is the charge for the
privilege of borrowing money, typically expressed as
annual percentage rate. Interest can also refer to the
amount of ownership a stockholder has in a
company, usually expressed as a percentage like
1.5%,or a decimal 0.015.
There are two main types of interest that can be
applied to loans: simple and compound. Simple
interest is a set rate on the principle originally lent to
the borrower that the borrower has to pay for the
ability to use the money. Compound interest is
interest on both the principle and the compounding
interest paid on that loan. The latter of the two types
of interest is the most common.
Interest rates have a huge efect on loans. In short,
loans with high interest rates have higher monthly
payments or take longer to pay of than loans with
low interest rates.
In mathematics, a geometric series, also known as a
geometric sequence, is a sequence of numbers
where each term after the frst is found by
multiplying the previous one by a fxed, non-zero
number called the common ratio.
Each credit card has a maximum limit. A maximum
limit of RM20000 means total money spent plus
interests incurred must not exceed this amount. This
can ensure that the credit card users will not spent
too much money by using credit card. Each credit
card user will choose a payment plan to repay the
debt every month.
Methodology
i=
Annual interest rate
12
Annual interest rate =18%
i = monthly interest rate
We use 18% divided 12 and get a answer 1.5% .1.5% is our monthly
interest rate.
Month,
n
Interest
due
(RM)
0
1
Payment,
R
(RM)
_
120
Paid to
principal
2
520
3
520
4
520
5
520
6
520
(RM)
(RM)
_
520
Balance,bn
_
400
Interest due = Balance,bn ×1.5%
Paid to principal = Payment,R - Interest due
8000
7600
Balance,bn = Balance,bn-1 - Paid to principal
bn =arn-1
a =8000
Write equation for b1,b2,b3,fnd the frequency and relationship
amog them by compare their diference and make a formula
for bn
For example: equation b1 is b1 =b0 – [ R-(b0)(i)]
Find the equation forb2 and b3 with equation b1.
make three table with three amont larger than 520 such as
600.800.1000 to fnd monthly balance until balance bn is
less than payment.The last month,n will be the month paid
of the payment
Interest due = Balance,bn ×1.5%
Paid to principal = Payment,R - Interest due
Balance,bn = Balance,bn-1 - Paid to principal
bn =arn-1
a =8000
Payment= 520,600,800,1000
When bn =0,
0= bn-1 – [ R-(bn-1)(i)]
b1 =b0 – [ R-(b0)(i)]
We can sub n=0 to fnd equation about b0
Use equation b1 =b0(1+i),b2,b3 and bn.Show bn = b0(1+i)
s
(1+i+ 100 )n-1
b2=b0(1+i)(1+i+ s )
100
b3=b0(1+i)(1+i+ s )2
100
Try to fnd relationship among b1, b2,b3 ,to form a equation for
bn
1
When s=1,sub s=1 into equation bn = b0(1+i)(1+i+ 100 )n-1
1
bn = b0(1+i)(1+i+ 100 )n-1
Then ,sub bn =16000
b0 =8000
i= 0.015
1
16000=8000(1+0.015)(1+0.015 + 100 )n-1
Solve the equation to fnd n
Result
monthly interest rate
i=
Annual interest rate
12
i = 18%
12
i = 1.5%
i=monthly interest rate =1.5%
Every mont ,we will be charge 1.5% of interest rate from our
balance.
This interest rate make us paid more mony then
beginning.Interest rate is also a way to let bank earn money.
When payment =RM520
Month, Interest Payment, Paid to Balance,
n
due
R
princip bn
al
(RM)
(RM)
(RM)
(RM)
0
1
2
3
4
5
6
7
_
120
114
107.91
101.73
95.45
89.09
_
520
520
520
520
520
520
520
82.62
8
9
10
520
76.06
69.4
520
520
62.64
11
520
55.78
12
13
48.
82
520
520
_
400
406
412.09
418.27
424.55
430.91
437.3
8
443.9
4
450.6
457.3
6
464.2
2
8000
7600
7194
6781.9
6363.6
5939.1
5508.2
5070.82
4626.88
4176.28
3718.92
3254.7
471.1
8 2783.52
478.2
5 2305.27
485.4
2 1819.85
492.7 1327.15
500.0
9 827.06
507.5
9 319.47
41.75
14
15
16
520
34.58
27.30
520
520
19.91
17
520
12.41
When payment =RM520,month balance paid of =17
At 17 th month,payment from RM 520 increase to RM
839.47.This make the payment faster paid end and save the
interest rate charge by the bank. At 17th month , balance is
fully paid of.
When payment =RM 600
Month,
n
Interest Payment,R Paid to
Balance,bn
due
principal
(RM)
0 _
_
1
120
2
112.8
3 105.49
4
98.07
5
90.55
6
82.9
7 75.15
8 67.28
9 59.29
10 51.18
11 42.94
12 34.59
(RM)
(RM)
_
600
600
600
600
600
600
600
600
600
600
600
600
480
487.2
494.51
501.93
509.45
517.1
524.85
532.72
540.71
548.82
557.06
565.41
(RM)
8000
7520
7032.8
6538.3
6036.4
5526.9
5009.8
4485.15
3952.43
3411.72
2862.9
2305.84
1740.43
13
14
26.11
17.50
600
600
573.89
582.5
1166.54
584.04
When payment =RM600,month balance paid of =14
At 14 th month,payment from RM 600 increase to RM
1184.04This make the payment faster paid end and save the
interest rate charge by the bank. At 14th month , balance is
fully paid of.
When payment =RM 650
Month,
n
Interest Payment,R Paid to
Balance,bn
due
principal
(RM)
(RM)
0 _
_
1
120
2 112.05
3 103.98
4 95.79
650
650
650
650
5
87.48
650
6
79.04
650
_
650
7
8
70.48
650
61.78
(RM)
530
537.95
546.02
554.2
1
562.5
2
570.9
6
579.5
2
588.2
2
(RM)
8000
7470
6932.05
6386.03
5831.8
2
5269.3
0
4698.3
4
4118.8
2
3530.6
0
9
650
52.96
10
650
44
11
650
34.91
12
650
25.69
13
650
16.32
597.0
4
2933.5
6
2327.5
6
1712.4
7
1088.1
6
606
615.0
9
624.3
1
633.6
8 454.48
When payment =RM650,month balance paid of =13
At 13 th month,payment from RM 650 increase to RM
1094.48 .This make the payment faster paid end and save
the interest rate charge by the bank.At 13 th month , balance
is fully paid of.
When payment =RM800
Month,n Interes
t due
(RM)
Payment,R Paid to Balance,bn
principal
(RM)
(RM)
(RM)
0 _
_
1
120
2
109.8
3
99.45
4
88.94
5
78.27
6
67.45
7 56.46
8 45.31
9 33.99
10
22.5
_
800
800
800
800
800
800
800
800
800
800
680
690.20
700.60
711.10
721.70
732.60
743.50
754.70
766.01
777.50
8000
7320
6629.80
5929.30
5218.20
4496.50
3763.90
3020.50
2265.80
1499.80
722.30
When payment =RM800,month balance paid of =10
At 10th month,payment from RM 800 increase to RM1522.30
This make the payment faster paid end and save the interest
rate charge by the bank. At 10th month , balance is fully paid
of.
When payment =RM1000
Month,n Interest Payment,R Paid to Balance,bn
due
principal
(RM)
(RM)
(RM)
(RM)
0 _
_
_
8000
1
120
1000
880
7120
2
106.8
1000
893.2
6226.8
3
93.4
1000
906.6
5320.2
4
79.8
1000
920.2
4400
5
66
1000
934
3466
6
51.99
1000 948.01
2517.99
7 37.77
1000
51.99
2466
8 36.99
1000 963.01 1502.99
9 22.54
1000 977.46
525.53
When payment =RM1000,month balance paid of =9
At 9 th month,payment from RM 1000 increase to
RM1525.53.This make the payment faster paid end and save
the interest rate charge by the bank. At 9 th month , balance
is fully paid of.
b1 =b0 – [ R-(b0)(i)] b1 = b0 – ( R – ib0 )
= ( 1 + i )b0 – R
b2 = b1 – ( R – ib1 )
= ( 1 + i )b1 – R
Substitute b1 into b2 , b2 = ( 1 + i ) [ ( 1 + i )b0 – R] – R
= ( 1 + i ) 2b0 - ( 1 + i )R – R
b3 = b2 – ( R – ib2 )
= ( 1 + i )b2 – R
Substitute b2 into b3 , b3 = ( 1 + i ) [ ( 1 + i )2b0 - ( 1 + i )R –
R]–R
= ( 1 + i ) 3b0 - ( 1 + i )2R - ( 1 +
i )R – R
After expressed b1 , b2 and b3 in term of R, i, n, and b0, I
observe relationship between them to form a formula for bn.
bn = b0( 1 + i )n - ( 1 + i )n-1R - ( 1 + i )n-2R - … - ( 1 + i )R – R
= b0( 1 + i )n– R [( 1 + i )n-1 + ( 1 + i )n-2 + … + ( 1 + i ) – 1]
= b0( 1 + i )n– R [1 + ( 1 + i ) + ( 1 + i )2 + ( 1 + i )3 + … +
( 1 + i )n-1]
Formula of sum:
a=1
, r=1+i
a(r n−1)
Sn = r−1
1[(1+i)−1]
= (1+i)−1
(1+i)n−1
=
i
Formula for bn
n−1]
bn = b0( 1 + i )n– R [(1+i)¿¿
¿
i
bn =balance in nth month
R =Payment
i = montly interest rate
n = month
bn = b0( 1 + i )n–
R [(1+i)¿¿ n−1]
¿
i
When payment = RM 520
n−1]
bn =8000 ( 1 + 0.015 )n– 520[(1+0.015)¿¿
¿
0.015
bn =0
8000 ( 1 + 0.015 )n–
520[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 520[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 520[( 1+ 0.015)¿¿ n−1] ¿ =0
n
120(1+0.015) – 520 ¿ ¿ =- 520
(1+0.015)n [120-520] = -520
n
(1+0.015) [(-400)] =-520
(1+0.015)n = 1.3
n log10 1.015 = log10 1.3
log 1.3
n= log 1.015
n =17.62
The month when balance is paid of while payment RM520 is
about 18 month
When payment = RM 600
bn =8000 ( 1 + 0.015 )n–
600[(1+0.015)¿¿ n−1]
¿
0.015
bn =0
n−1]
8000 ( 1 + 0.015 )n– 600[(1+0.015)¿¿
¿ =0
0.015
120(1+0.015)n – 600[(1+0.015)¿¿ n−1]
¿ =0
0.015
n
120(1+0.015) – 600 [(1+ 0.015)¿¿ n−1]¿ =0
120(1+0.015)n – 600 ¿ ¿ =-600
n
(1+0.015) [120-600] = -600
(1+0.015)n[(-480)] =-600
n
(1+0.015)
= 1.25
n log10 1.015 = log10 1.25
log 1.25
n= log 1.015
n =14.99
The month when balance is paid of while payment RM520 is
about 15 month
When payment = RM 650
n−1]
bn =8000 ( 1 + 0.015 )n– 650[(1+0.015)¿¿
¿
0.015
bn =0
8000 ( 1 + 0.015 )n–
650[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 650[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 65 0 [(1+ 0.015)¿¿ n−1]¿ =0
n
120(1+0.015) – 650 ¿ ¿ =- 650
(1+0.015)n [120-650] = -650
n
(1+0.015) [(-530)] =-650
(1+0.015)n = 1.23
n log10 1.015 = log10 1.23
log 1.23
n= log 1.015
n =13.90
The month when balance is paid of while payment RM520 is
about 14 month
When payment = RM 800
bn =8000 ( 1 + 0.015 )n–
800[(1+0.015)¿¿ n−1]
¿
0.015
bn =0
n−1]
8000 ( 1 + 0.015 )n– 800[(1+0.015)¿¿
¿ =0
0.015
120(1+0.015)n – 520[(1+0.015)¿¿ n−1]
¿ =0
0.015
n
120(1+0.015) – 800 [(1+ 0.015)¿ ¿ n−1]¿ =0
120(1+0.015)n – 800 ¿ ¿ =- 800
n
(1+0.015) [120-800] = -800
(1+0.015)n[(-680)] =-800
n
(1+0.015)
= 1.18
n log10 1.015 = log10 1.18
log 1.18
n= log 1.015
n =11.11
The month when balance is paid of while payment RM520 is
about 11 month
When payment = RM 1000
n−1]
bn =8000 ( 1 + 0.015 )n– 1000[(1+0.015)¿¿
¿
0.015
bn =0
8000 ( 1 + 0.015 )n–
1000[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 1000[(1+0.015)¿¿ n−1]
¿ =0
0.015
120(1+0.015)n – 1000[(1+ 0.015)¿¿ n−1]¿ =0
n
120(1+0.015) – 1000 ¿ ¿ =- 1000
(1+0.015)n [120-1000] = -1000
n
(1+0.015) [(-880)] =-1000
(1+0.015)n = 1.14
n log10 1.015 = log10 1.14
log 1.14
n= log 1.015
n =8.8
The month when balance is paid of while payment RM520 is
about 9 month
bn = b0( 1 + i )n-
R [(1+i)¿¿ n−1]
¿
i
If bn = 0 ,
n−1]
b0( 1 + i )n - R [(1+i)¿¿
¿ =0
i
b 0(1+i)n (i)−R [(1+i)¿¿ n−1]
¿ =0
i
b0( 1 + i )n(i) – R(1 + i)n + R = 0
b0( 1 + i )n(i) = R( 1 + i )n – R
b0 =
R [(1+i)¿¿ n−1]
¿
i(1+i)n
R
R
b0 = i - i(1+i)n
b0 =R(i)-1- R[i(1+i)n]-1
If bn = 0 , there has an expression for b0 in terms of i, n and
R.
b1=b0(1+i)
b2=b0(1+i)(1+i+ s )
100
b3=b0(1+i)(1+i+ s )2
100
b4=b0(1+i)(1+i+ s )3
100
b5=b0(1+i)(1+i+ s )4
100
b6=b0(1+i)(1+i+ s )5
100
bn=b0(1+i)(1+i+ s )n-1
100
When S=1
1
bn = b0(1+i)(1+i+ 100 )n-1
1
16000=8000(1+0.015)(1+0.015 + 100 )n-1
41
400
( 4 o )n-1 = 203
41
400
(n-1) log10 ( 4 o )= log10 203
(n-1) =27.5
n=28.4589
n=28.46
When S=2
2
bn = b0(1+i)(1+i+ 100 )n-1
2
16000=8000(1+0.015)(1+0.015 + 100 )n-1
(
207
200
400
)n-1 = 203
207
400
(n-1) log10 ( 200 )= log10 203
(n-1) =19.933
n= 20.933
When 1% is charge,Current payment is doubled after 28
month.
When 2% is charge,Current payment is doubled after 20
month.
Conclusion
In conclusion,the higher the montly
payment,the faster the time of balance paid
of,the decrease the month of the balance paid
of.If the man began and continue paid of the
current balance ,the balance can paid of
faster.If the man decide not to pay at all,the
balance will become more and have to pay
more at future.When the number of s%
increase , month when current payment double
increase.
Furthermore, interest rate will increase when
we need longer time to pay of our
payment.Balance we pay will decrease while
intrest increase through geometry series .We
need to used the shorter time we can to pay of
our payment to reduce the charge of
interest.The higher the interest rate ,the higher
the debt.The shorter the month to paid of the
debt,the less the total amount we paid .As a
consumer,we need to be wise when paying our
debt.We need to have a good planning to pay
of the debt to avoid debt become more until
we cannot repay.