repayments (1 Viewer)

[mizz_smee]

i can't get the logic behind repayments
especially when it is superannuation
can someone please tell me how to apply
a geometric series or whatever u use
you can use any question u wish
just explain it don't just show the working
if someone could help that would be appriciated
thanks.

 

[Jago]

you're asking maths people to explain using words? HAH!

 

[velox]

Ok Here goes

A person invests $2000 at the beginning of each year in a super fund. Assuming interest is paid at the rate of 9% per annum, find out how much the inverstment is worth at the end of 40years.

We'll see what the 1st $2000 that was invested, after 40 years:
Year One $2000 invested = 2000(1+0.09)<SUP>40</SUP>
Year Two $2000 invested = 2000(1+0.09)<SUP>39</SUP>
Year Three $2000 invested= 2000(1+0.09)<SUP>38</SUP>

As you can see the term (n) is getting lower by one each time, as he deposits $2000 at the beginning of each year. Like the first $2000 will be in their for the full 40 years, whereas the 2nd $2000 that he deposits will only be there for 39 years etc.
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Year Forty $2000 invested = 2000(1+0.09)<SUP>1</SUP>
(as its only in the super fund for a year)

The sum of these 40 equations = 2000(1.09)<SUP>1</SUP>+ ............+2000(1.09)<SUP>39</SUP>+2000(1.09)<SUP>40</SUP>

Which after factorising will be
2000(1.09<SUP>1</SUP>+1.09<SUP>2</SUP>+1.09<SUP>3</SUP>+.............+1.09<SUP>40</SUP>)
As it is a geometric series increasing by x1.09 each time, we can use the sum formula:

s<SUB>n</SUB> = (a(1-r<SUP>n</SUP>))/(1-r)

Where a= 1.09
r = 1.09
n = 40

s<SUB>40</SUB> = (1.09(1-1.09<SUP>40</SUP>))/(1-1.09)


We sum it and then get the investment to be worth $736 583.73 (to the nearest cent) after the 40 years is up.

 

[Tommy_Lamp]

Ok heres a question in my textbook on superannuation that I've done and my answer is different to back of the book.

A woman invests $600 at the beginning of each year into a super fund for a period of 10 years. Interest is paid at the rate of 7% p.a. and is compounded annually. Find how much her investment is worth at the end of the 10 year period.

I wont go through my entire working, but my last line came out as:
600 x ( 1.08 (1 - 1.08^20 ) ) / (1 - 1.08)
= $29,653.75

The answer in the book is $8870

Thanks

 

[velox]

Tommy Lamp- That's because you have used 8% as the rate in the sum formula instead of 7%. Try it with 1.07 instead of 1.08 and it works out fine.

 

[Tommy_Lamp]

Oh shit ><
Pre-exam stress...must relax

 

[velox]

And another thing i forgot to mention, is that n = 10, not 20. That also messed up the answer.

 

[Purp|e]

ever since the trial when i had no idea what i was doing for one of these questions and i got 7/7 for it they have been my favourite

 

[Seraph]

well at least the question wrx posted said that its deposited at the beggining of the year

when its not

oi.... thats where it gets tricky

 

[Tommy_Lamp]

Ok I've done 5 other questions and its all good, thanks wrx

Seraph, are you talking about compunded monthly etc.?
Or when you have to add another year to the total?

 

[Seraph]

add another year .. to the total or you haev one too many years.. those type of questions

compounded monthly, half yearly etc blah thats not too bad

divide by 4 or divide by 2 for itnerest rate and years

 

[velox]

When it is at the end of the year its still very simple.

Here is the same question as above, but with the $2000 being deposited at the end of each year.

We'll see what the 1st $2000 that was invested, after 40 years:
Year One $2000 invested = 2000(1+0.09)<SUP>39</SUP>
Year Two $2000 invested = 2000(1+0.09)<SUP>38</SUP>
Year Three $2000 invested= 2000(1+0.09)<SUP>37</SUP>

As you can see the term (n) is getting lower by one each time. The 1st $2000 is in there for 39 years (not 40) as it is deposited at the end of each year.
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Year Forty $2000 invested = 2000(1+0.09)<SUP>0</SUP>
(as he deposits the $2000 at the end of each year, the last $2000 will get no interest)

The sum of these 40 equations = 2000 + 2000(1.09)<SUP>1</SUP>+ ............+2000(1.09)<SUP>38</SUP>+2000(1.09)<SUP>39</SUP>

Which after factorising will be
2000(1.09<SUP>1</SUP>+1.09<SUP>2</SUP>+1.09<SUP>3</SUP>+.............+1.09<SUP>39</SUP>)
As it is a geometric series increasing by x1.09 each time, we can use the sum formula:

s<SUB>n</SUB> = (a(1-r<SUP>n</SUP>))/(1-r)

Where a= 1.09
r = 1.09
n = 39 (as the 40th term is just 2000, so you just add it on)

s<SUB>39</SUB> =2000+ ((1.09(1-1.09<SUP>39</SUP>))/(1-1.09))


We sum it and then get the investment to be worth $675 765 (to the nearest dollar) after the 40 years are up.

Its pretty amazing how much the difference between depositing the money at the start of the year and depositing the money at the end of year makes. Its over $60 000 difference over 40 years!

 

[Tommy_Lamp]

So if it is being deposited at the end of the year, then you take one year off the total because no interest is payed on the last year?

 

[velox]

Yes, because its asking what the investment is worth at the end of 40 years. So the last $2000 will get no interest.

So you take one off the 'n' and then add the amount being invested at the end, after you have summed the rest of the terms.

But remember to add the orignal amount, a lot of people forgot.

 

[PC]

Have a look at hsc.csu.edu.au/maths and follow the links to the sections on applications of series. There's a good PDF there with fully worked solutions on Time Repayments and Superannuation questions.

 

[Seraph]

hmm so interest is ALWAYS charged at the end of each year?

 

[velox]

hmm so interest is ALWAYS charged at the end of each year?

Im pretty sure that if its says compounded yearly it would be at the end of the year. And if it said compounded monthly the interest will be charged at the end of each month.

So, yes.

 

[Seraph]

fair enough

seems a bit more clearer to me now

 

[Tommy_Lamp]

PC: I'm looking at that right now
http://hsc.csu.edu.au/maths/mathematics/series_apps/time_payments/time.pdf

 

[mizz_smee]

thank you much clearer now
just gotta remember it for the exam now