Geometric sequences

  • #1
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Homework Statement


A mortgage is taken out for £80,000. It is to be paid by annual instalments of 5000with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.


Homework Equations


Sn=a(r^n-1)/r-1


The Attempt at a Solution



I have no idea here can someone please point me in the right direction, thanks!
 

Answers and Replies

  • #2

Homework Statement


A mortgage is taken out for £80,000. It is to be paid by annual instalments of 5000with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.


Homework Equations


Sn=a(r^n-1)/r-1


The Attempt at a Solution



I have no idea here can someone please point me in the right direction, thanks!

Since you want a pointer in the right direction, start with your 80K subtract the payment 5K...then add to what remains 4% interest, then subtract 5K again and add to what remains 4% interest, then subtract...and so on. (so long as you don't actually calculate anything here you should see that you'll be summing up terms in a geometric sequence)
Clearly if you keep repeating this process the debt will become zero. the n in your relevant equation is the number of times interest gets added.
For what value of n will it be true that your above summation will be equal to zero?
 
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  • #4
I'm getting just over 12 years, is that correct?
 
  • #5
it dosen't sound right
 
  • #6
with no interest, paying 5000/year. it would take 80000/5000 = 16 years to pay it off. it's going to be more than that with interest
 
  • #7
I'm getting just over 12 years, is that correct?

what expression did you form to get that?
furthermore, shouldn't your sum equation be of the form: Sn = (r^(n+1)-1)/(r-1)?

Hmm..am I right in saying that you got 12 years by the following process:
1.a): 80000-5000
1.b): (1.a) - 0.04*(1.a)
1.c): (1.b) - 5000
1.d): (1.c) - 0.04*(1.c)...and so on?
 
Last edited:

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