Help! Loan Repayments Question (1 Viewer)

tonyharrison

Member
Joined
Mar 17, 2009
Messages
366
Gender
Female
HSC
2009
If you know:
the monthly repayment ($2800)
the montly interest (1.006)
the total amount needed to be repaid ($200 000)

How do you find how many months it will take?

I don't need to know the numbers, just a quick overview of how to do it.
If anyone could help, it would be much appreciated. Thanks!
 

raniaaa

:)
Joined
Mar 21, 2008
Messages
480
Gender
Female
HSC
2009
P= $200 000
R= 1.006
M= $2 800
n= ?


A1 = PR^1 - M
A2 = (A1)R - M
= PR^2 - MR - M
= PR^2 - M [ 1+ R]
A3 = PR^3 - M [ 1 + R + R^2]
.
.
.
.
An = PR^n - M [ 1 + R + R^2 ..... + R^n ]
but An = 0 (since loan has been paid off)
PR^n = M [ 1 + R + R^2 ..... + R^n ]
PR^n = M [ 1( R^n - 1) / R - 1 ] .... (i.e. sum of the series)
PR^n = M/R-1 (R^n) - M/R-1
sub values in:
200 000 (1.006)^n = 2 800/0.006 (1.006)^n - 2800/0.006
200 000 (1.006)^n = 466 666.67 (1.006)^n - 466 666.67
266 666.67 (1.006)^n = 466 666.67
(1.006)^n = 1.75
take logs of both sides
n ln 1.006 = ln 1.75
n = ln 1.75 / 1.006
answer in months


=)
 
Last edited:

chopstick

New Member
Joined
Sep 10, 2008
Messages
28
Gender
Male
HSC
N/A
you should come up with equations like A=200000x1.006^n--2800(1+1.006....1.006^(n-1)) and then A=0 at the end of last month and use the geo sum, so 0=200000x1.006^n-2800(1x(1.006^n -1)/1.006-1) times all by 0.006 0=200000x0.006x1.006^n -- 2800x1.006^n + 2800 0=(1200--2800)x1.006^n +2800 1600/2800=1.006^n then log both side log(1600/2800)=log(1.006^n) =n log1.006 divide both side to find n, and n must be a whole number. i dont know how to give a overview so... i not sure there are any mistake or not hope you can understand what i wrote. why is the word all stick together!!!!
 
Last edited:

Cloesd

Member
Joined
Nov 4, 2008
Messages
156
Gender
Male
HSC
2009
If you know:
the monthly repayment ($2800)
the montly interest (1.006)
the total amount needed to be repaid ($200 000)

How do you find how many months it will take?

I don't need to know the numbers, just a quick overview of how to do it.
If anyone could help, it would be much appreciated. Thanks!
I'v always had a problem with these types of questions but never got the chance to experiment to figure it out.

My problem is: Does it matter if the interest is applied first, and THEN the repayment is made?
 

sam64mcd

New Member
Joined
Mar 22, 2009
Messages
16
Location
Central Coast
Gender
Female
HSC
2009
Yeah, interest is applied to the 'A' or amount before the monthly repayment is decreased from the total 'A' for the next installment after it. That way, when you derive an equation for 'An', the interest on 'M' becomes a geometric series. It also makes sense that interest would be added on to the principle before the repayment is made. Hope that answered your question..
 

Cloesd

Member
Joined
Nov 4, 2008
Messages
156
Gender
Male
HSC
2009
Yeah, interest is applied to the 'A' or amount before the monthly repayment is decreased from the total 'A' for the next installment after it. That way, when you derive an equation for 'An', the interest on 'M' becomes a geometric series. It also makes sense that interest would be added on to the principle before the repayment is made. Hope that answered your question..
Awesome.
 

tonyharrison

Member
Joined
Mar 17, 2009
Messages
366
Gender
Female
HSC
2009
you should come up with equations like A=200000x1.006^n--2800(1+1.006....1.006^(n-1)) and then A=0 at the end of last month and use the geo sum, so 0=200000x1.006^n-2800(1x(1.006^n -1)/1.006-1) times all by 0.006 0=200000x0.006x1.006^n -- 2800x1.006^n + 2800 0=(1200--2800)x1.006^n +2800 1600/2800=1.006^n then log both side log(1600/2800)=log(1.006^n) =n log1.006 divide both side to find n, and n must be a whole number. i dont know how to give a overview so... i not sure there are any mistake or not hope you can understand what i wrote. why is the word all stick together!!!!
About the words "Stick"ing together, check my thread at Writing Appearing in a Block!
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top