C2 - Geometric Series

A mortgage is taken out for 80000 pounds. It is to be paid by annual instalments of 5000 pounds with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4 per cent is then charged on any outstanding debt. Find the total time taken to pay off the mortgage-

I am stuck on this one, can anyone help?

thanks

Reply 1

steve2005

17

shivmani

A mortgage is taken out for 80000 pounds. It is to be paid by annual instalments of 5000 pounds with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4 per cent is then charged on any outstanding debt. Find the total time taken to pay off the mortgage-

I am stuck on this one, can anyone help?

thanks

I have started but not yet got to the sequence

Picture 128.png24.9KB

Reply 2

DFranklin

18

Unparseable latex formula:

a_0 = 80000. \\a_1 = (80000-5000)\times 1.04. \\[br]a_2 = ((80000-5000)\times 1.04)-5000)\times 1.04 = 80000\times (1.04)^2 - 5000(1.04+1.04^2)\\[br]a_n = 80000 \times (1.04)^n -5000\sum_1^n (1.04)^k \\[br] = 80000\times (1.04)^n - 5000 \times 1.04 \frac{1.04^n-1}{1.04-1} \\[br] = 80000\times (1.04)^n - 130000(1.04^n-1) \\[br] = 130000 - (1.04)^n\times (130000 - 80000) \\[br] = 130000 - 50000 \times (1.04)^n

Sequence goes negative when

1.04^n > \frac{130000}{50000} \implies n \log (1.04) > \log(13/5) \implies n > 24.36 \implies n = 25

.

Reply 3

shivmani

OP

thanks, but that's exactly where i got to. The answer is 25 years . The hint : find an expression for the debt remianing after n years and solve using the fact that if it is paid off, the debt = 0.

Reply 4

shivmani

OP

thanks so much, I will have to go over that slowly to understand it.

Reply 5

steve2005

17

shivmani

thanks so much, I will have to go over that slowly to understand it.

Beaten, but I am pleased that I get the same answer as DFranklin.

Picture 129.png75.0KB

Reply 6

shivmani

OP

I still dont understand where the 130000 come from?

Reply 7

DFranklin

18

5000 x 1.04 / (1.04-1) = 5200 / .04 = 130000.

Reply 8

shivmani

OP

ofcourse, how silly of me not to see that. thank you

Reply 9

Theorist

10

DFranklin

Unparseable latex formula:

a_0 = 80000. \\a_1 = (80000-5000)\times 1.04. \\[br]\mathbf{a_2 = ((80000-5000)\times 1.04)-5000)\times 1.04 = 80000\times (1.04)^2 - 5000(1.04+1.04^2)}\\[br]\mathbf{a_n = 80000 \times (1.04)^n -5000\sum_1^n (1.04)^k} \\[br] = 80000\times (1.04)^n - 5000 \times 1.04 \frac{1.04^n-1}{1.04-1} \\[br] = 80000\times (1.04)^n - 130000(1.04^n-1) \\[br] = 130000 - (1.04)^n\times (130000 - 80000) \\[br] = 130000 - 50000 \times (1.04)^n

Sequence goes negative when

1.04^n > \frac{130000}{50000} \implies n \log (1.04) > \log(13/5) \implies n > 24.36 \implies n = 25

.

I'm sorry but I don't understand how you got the two lines, marked in bold. Please could you go over these in a little more detail?

Reply 10

DFranklin

18

Dharma

I'm sorry but I don't understand how you got the two lines, marked in bold. Please could you go over these in a little more detail?

All I've done in the first line is use

a_2 = (a_1 - 5000) \times 1.04

(since to get the sum after 2 years we take the sum after 1 year, subtract the repayment of 5000 and multiply the balance by 1.04) and then multiplied it all out and rearranged.

The

a_n

line is a bit more of a jump although I would expect an examiner to follow it. Basically after each year, we end up multiplying the 80000(1.04)^n term by 1.04 to get 80000(1.04)^(n+1). We also multiply the -5000(1.04+1.04^2+...1.04^n) term by 1.04 and since we subtract off another 5000 x 1.04 term at the same time we end up with -5000(1.04+1.04^2+...1.04^(n+1). You might find it easier to see what's going on if you calculate a few more terms by hand.

If you want to get formal, I suppose you could prove it by induction, but I don't see the point for something so obvious.

Reply 11

Roshniroxy

14

Original post by DFranklin

Unparseable latex formula:

a_0 = 80000. \\a_1 = (80000-5000)\times 1.04. \\[br]a_2 = ((80000-5000)\times 1.04)-5000)\times 1.04 = 80000\times (1.04)^2 - 5000(1.04+1.04^2)\\[br]a_n = 80000 \times (1.04)^n -5000\sum_1^n (1.04)^k \\[br] = 80000\times (1.04)^n - 5000 \times 1.04 \frac{1.04^n-1}{1.04-1} \\[br] = 80000\times (1.04)^n - 130000(1.04^n-1) \\[br] = 130000 - (1.04)^n\times (130000 - 80000) \\[br] = 130000 - 50000 \times (1.04)^n

Sequence goes negative when

1.04^n > \frac{130000}{50000} \implies n \log (1.04) > \log(13/5) \implies n > 24.36 \implies n = 25

.

sorry i realise this is four years late but I'm attempting to do this question and I just can't figure out why when you do the sum of n years why the value for 'a'= 1.04, wouldn't it be £80,000

C2 - Geometric Series

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