# geometrical progression

• Sep 8th 2009, 10:33 PM
furor celtica
geometrical progression
im not sure where to post this
so
a bank loan of 500 pounds is arranged to be repaid in two years by equal monthly instalments. Interest, CALCULATED MONTHLY, is charged at 11% per annum on the remaining debt. Calculate the monthly repayment if the first repayment is to be made one month after the loan is granted.

I've been working for almost two hours on this problem, ending up with the equation (500-m)(1211/1200)^24 - (1200m/11) ((1211/1200)^24 -1)
m being the equal monthly repayment
from there i ended up with m= 21.something, apparently not far from the truth but nevertheless incorrect; can somebody show me how they would do this?
• Sep 8th 2009, 11:25 PM
Wilmer
P = Ai / [1 - 1/(1 + i)^n]

A = Amount borrowed (500)
i = interest per month (.11/12)
n = number of payments (24)
• Sep 8th 2009, 11:29 PM
furor celtica
thanks but i can't use this thing, i need to use geometric progression methods
• Sep 9th 2009, 01:11 AM
furor celtica
please i really need help here, ive posted this cos ive tried everything already
• Sep 9th 2009, 04:18 AM
furor celtica
ok so i got 23.29 when my textbook says 23.31; is this a negligible difference or a symptom of a BIG problem? this is my work
ahem
S1= (500-m monthly payment) x (1211/1200)
S2= (S1 -m) x (1211/1200)
= (500-m)(1211/1200)^2 - (1211/1200)m
and so on
so i take as general equation for finding m
0= (500-m)(1211/1200)^24 - m[(1211/1200)^23 -1]/[(1211/1200)-1]

from there i get 23.29 pounds
is there any fault in my working?
• Sep 9th 2009, 04:47 AM
Hello furor celtica
Quote:

Originally Posted by furor celtica
ok so i got 23.29 when my textbook says 23.31; is this a negligible difference or a symptom of a BIG problem? this is my work
ahem
S1= (500-m monthly payment) x (1211/1200)
S2= (S1 -m) x (1211/1200)
= (500-m)(1211/1200)^2 - (1211/1200)m
and so on
so i take as general equation for finding m
0= (500-m)(1211/1200)^24 - m[(1211/1200)^23 -1]/[(1211/1200)-1]

from there i get 23.29 pounds
is there any fault in my working?

I make it £23.30 to the nearest penny, so I reckon you're about OK. I don't know that I really understand your working, though. Here's how I did it:

At the end of month 1, the amount owing = $500\times\frac{1211}{1200}-m$

At the end of month 2, when we simplify the expression, the amount owing = $500\times\Big(\frac{1211}{1200}\Big)^2-\Big(m+m\times\frac{1211}{1200}\Big)$

At the end of month 3, the amount owing = $500\times\Big(\frac{1211}{1200}\Big)^3-\Big(m+m\times\frac{1211}{1200}+m\times\Big(\frac{ 1211}{1200}\Big)^2\Big)$

... and so on.

The expression in the bracket is a GP, first term $m$, common ratio $\frac{1211}{1200}$, sum to 24 terms = $m\frac{\Big(\dfrac{1211}{1200}\Big)^{24}-1}{\dfrac{1211}{1200}-1}$

If the sum owing at the end of month 24 is zero, then:

$500\times\Big(\frac{1211}{1200}\Big)^{24} - m\frac{\Big(\dfrac{1211}{1200}\Big)^{24}-1}{\dfrac{1211}{1200}-1}=0$

...which you then solve, giving £23.30 to the nearest penny.

• Sep 9th 2009, 04:53 AM
furor celtica
are you sure its not to 23 terms?(Wondering)
what do you not understand in my working?
• Sep 9th 2009, 06:04 AM
Wilmer

Plus he IS using the equivalent of 23 terms:
the divisor (1211/1200 - 1) can be moved to the numerator as (1211/1200 - 1)^(-1)
• Sep 9th 2009, 09:52 PM
Hello furor celtica
Quote:

Originally Posted by furor celtica
are you sure its not to 23 terms?(Wondering)
what do you not understand in my working?