Author Topic: Help with a loan question. (Read 2384 times) Tweet Share

nickglyn · « **on:** October 18, 2016, 11:50:34 am »

Hey guys. I just need help with part ii) of question b. I know the steps of what to do but my algebra isn't good with these types of questions. Could you show me a worked solution please? Thank you!

jakesilove · « **Reply #1 on:** October 18, 2016, 12:03:17 pm »

Quote from: nickglyn on October 18, 2016, 11:50:34 am

Hey guys. I just need help with part ii) of question b. I know the steps of what to do but my algebra isn't good with these types of questions. Could you show me a worked solution please? Thank you!

Hey! So it's clear that

$A_n=200,000r^n-2,800(1+r+r^2+...+r^{n-1})$

We want to find the value for n that will completely pay off this loan. As A is the amount due after n repayments, we can just set A to zero! This gives us

$0=200,000r^n-2,800(1+r+r^2+...+r^{n-1})$

Now, we need to sum up the geometric series. We do this using the relationship

$S_n=\frac{a(1-r^n)}{1-r}$

I always struggle to figure out how many terms there are. In this case, we know that the first term is 1, the second term is r, the third term is r^2 and so on. Therefore, whatever the power of r is, the term is one more than that! So, since the final term is r^(n-1), there will be n terms.

$S_n=\frac{1(1-(1.006)^n)}{1-(1.006)}=$

$0=200,000(1.006)^n-2,800*(-166.6667)(1-(1.006)^n)$

Expanding gives us

$0=200,000(1.006)^n+466667-466667(1.006)^n$
$266666(1.006)^n=466667$
$(1.006)^n=1.75$
$log_{1.006}1.75=n=93.5 months$

Hope I didn't screw up anything there! Let me know if I can clarify anything

nickglyn · « **Reply #2 on:** October 18, 2016, 03:32:19 pm »

Hey so I just need a few things clarified. Why did you use the geometric sum equation for r<1? Isn't our 1.006? Also I'm not sure where you got -166.66667 from :/ sorry hahah

jakesilove · « **Reply #3 on:** October 18, 2016, 03:34:31 pm »

Quote from: nickglyn on October 18, 2016, 03:32:19 pm

Hey so I just need a few things clarified. Why did you use the geometric sum equation for r<1? Isn't our 1.006? Also I'm not sure where you got -166.66667 from :/ sorry hahah

I didn't use the geometric sum for r<1? That's just the standard geometric sum formula. If you mean why I used 1-r^n instead of r^n-1, the formulas are exactly the same either way, one may just be more convenient depending on the value for r.

The -166.66667 was just simplifying the 1/(1-1.006) in the geometric sum equation

RuiAce · « **Reply #4 on:** October 18, 2016, 03:36:24 pm »

nickglyn · « **Reply #5 on:** October 18, 2016, 03:39:41 pm »

OHHHHH okay, thanks heaps!! I'm sorry if I came across as rude I didn't intend on that I was just curious!

jakesilove · « **Reply #6 on:** October 18, 2016, 03:45:26 pm »

Quote from: nickglyn on October 18, 2016, 03:39:41 pm

OHHHHH okay, thanks heaps!! I'm sorry if I came across as rude I didn't intend on that I was just curious!

Aha no you didn't come across like that! All good

Search the forums now!

We have moved!

Author Topic: Help with a loan question. (Read 2384 times) Tweet Share

nickglyn

Help with a loan question.

jakesilove

Re: Help with a loan question.

nickglyn

Re: Help with a loan question.

jakesilove

Re: Help with a loan question.

RuiAce

Re: Help with a loan question.

nickglyn

Re: Help with a loan question.

jakesilove

Re: Help with a loan question.

Recent Posts

User:
Password: