# Loan Problem Help

• Feb 8th 2010, 03:24 PM
princessamme@hotmail.com
Loan Problem Help
An investor raises £100m in order to buy a football club. The terms of the repayment schedule specify that the loan must be paid in 12 six monthly instalments, the first falling due six months from now, with each instalment being 20% larger than the first. Thus the scond payment is 1.2 times the first and the third is 1.2 times the second ect. Interest is charged bi-annually at rate r2.

If Y is the value of the first payment, show that, (see formula below)

Any help on this would be greatly appreciated
http://texify.com/?$100=Y%20%5Ctimes...%20-%200.2%7D$
• Feb 9th 2010, 08:22 AM
Silver
I NEED AN ANSWER TO THIS PLEASE!
it is driving me mad, I have used the idea of partial sum of geometric series but whatever i do, i dont get what is in the brackets.
Any suggestions, even tips on what to try when proving a formula would be amazing please, thank you!

Btw, hello emma! :D
• Feb 9th 2010, 12:38 PM
princessamme@hotmail.com
Hey sophia, yeah what i've done is ive said its gonna be

100 = Y{ 1 + (1.2)^1(1+r2/2)^-2.....+(1.2)^11(1+r2/2)^-12

then tried to put it into the formula of (1+r)^n/r but it's not giving me the right answer how are u doin with it?
• Feb 9th 2010, 12:45 PM
Silver
What i did is nothing like that,
I tried to figure out what r is first, which is $1.2*(1+r)$
well thats what i kept getting, and everyone else also seems to be getting that too

if i DID get $r = 1.2/(1+r)$ then i suppose id be able to prove whats asked in the question...

but like i said, it makes no sense how i keep getting $1.2*(1+r)$
i am very confused atm anyway, have you done question 6 by any chance?

x
• Feb 9th 2010, 12:47 PM
Silver
how is it that in your equation (1+r) is raised to negative powers? :/
• Feb 9th 2010, 01:12 PM
princessamme@hotmail.com
I was trying to follow what he'd done in a similar less complicated problem but it doesn't seem to be workin anyway (Headbang). No can't do question 6 either, im getting really frustrated with the whole thing. I'm gonna come in early tomorrow and try and work throught it some more i think (Worried)
• Feb 9th 2010, 01:13 PM
Silver
And the funny thing is, this assignment is only 3% of the WHOLE thing, and thats like, what, an extra question in the exam?

lol -_- its so annoying
• Feb 9th 2010, 01:16 PM
Silver
Oh what example is it btw? So i can go through it and try to understand why its raised to -ive powers please?
thank you

I kind of understand 6, and try to do it but it gets so messy, ill show you what ive done tomorrow before killians (or in numerical methods if youre in my tutorial group)
• Feb 9th 2010, 01:30 PM
princessamme@hotmail.com
Ive been using question 1.12, 1.14 and 1.13 and trying to adapt them. I have no idea if it's right or not. But i don't know what to do with the 1.2 that's what's confusing me. Thanks that'd be appreciated :D
• Feb 9th 2010, 01:38 PM
Silver
ah yes thank you! ill have to look at them
i dont want to give up on 6 yet but i just realised ive being very blind and putting wrong numbers in
meh ill start AGAIN (Thinking)
• Feb 9th 2010, 04:08 PM
Robb
Hello,

You were on the right track here..

Quote:

Originally Posted by princessamme@hotmail.com
100 = Y{ 1 + (1.2)^1(1+r2/2)^-2.....+(1.2)^11(1+r2/2)^-12

From this,
$Y\cdot (1.2)^0 \cdot \left( \frac{1}{1+r/2}\right)^1+Y\cdot (1.2)^1 \cdot \left( \frac{1}{1+r/2}\right)^2+...+Y\cdot (1.2)^11 \cdot \left( \frac{1}{1+r/2}\right)^{12}$
This can be manipulated into a geometric progression;
$\frac{1}{1+r/2}\cdot Y\cdot \left( \frac{1.2}{1+r/2}\right)^0 + \frac{1}{1+r/2}\cdot Y\cdot \left( \frac{1.2}{1+r/2}\right)^1+...+\frac{1}{1+r/2}\cdot Y\cdot \left( \frac{1.2}{1+r/2}\right)^{11}$

Which is equal to;
$
\frac{1}{1+r/2}\cdot Y \cdot \frac{\left( \frac{1.2}{1+r/2} \right)^{12}-1}{\frac{1.2}{1+r/2}-1}
$

So multiplying the bottom of the fractions,
$
Y \cdot \frac{\left( \frac{1.2}{1+r/2} \right)^{12}-1}{.2-r/2}
$

So by multiplying top and bottom by -1, is equal to;
$
Y \cdot \frac{1-\left( \frac{1.2}{1+r/2} \right)^{12}}{r/2-.2}
$
• Feb 9th 2010, 09:49 PM
Wilmer
The general formula to calculate 1st payment is:

p = a(u - v) / [1 - {(1 + v)/(1 + u)}^n]

p = 1st payment
a = amount borrowed
u = periodic interest rate
v = percentage (payment increase)
n = number of payments

The formula can be easily rearranged in terms of a:

a = p [1 - {(1 + v)/(1 + u)}^n] / (u - v)