C2 geometric series

A mortage is taken for £80,000 .It is paid by annaul instalments of £5,000 with the first payment being made at the end of the first year that the mortage was taken out. Interest of 4% is then charged on the out standing debt. Find the the the total time taken to pay off the mortage.

Note: Find an experession for the debt remaining after n years and slove using the fact that if it is paid off, the debt = 0.

thanks for the help in advance. Can you explian how you came to finding the experession.

Reply 1

Jonny W

8

After 1 year,

debt
= (80000 - 5000)*1.04
= 80000*1.04 - 5000*1.04

After 2 years,

debt
= (debt after 1 year - 5000)*1.04
= ((80000 - 5000)*1.04 - 5000)*1.04
= 80000*1.04^2 - 5000*(1.04 + 1.04^2)

After 3 years,

debt
= (debt after 2 years - 5000)*1.04
= (80000*1.04^2 - 5000*(1 + 1.04 + 1.04^2))*1.04
= 80000*1.04^3 - 5000*(1.04 + 1.04^2 + 1.04^3)

After n years,

debt
= 80000*1.04^n - 5000*(1.04 + 1.04^2 + ... + 1.04^n)
= 80000*1.04^n - 5000*1.04(1 - 1.04^n)/(1 - 1.04)
= 130000 - 50000(1.04^n)

Solving 130000 - 50000(1.04^n) = 0 gives n = 24.36, so the mortgage is paid off (with some money to spare) after 25 years.

Reply 2

ScipioAfricanus

OP

Thanks for all the help. I only understand the first part and some of the second. The answer is right as it's the same as the on in the back of the book. What i dont really undertand is how you arrived at 130000 and 50000.

Reply 3

Jonny W

8

80000*1.04^n - 5000*1.04(1 - 1.04^n)/(1 - 1.04)
= 80000*1.04^n + 130000(1 - 1.04^n)
= 130000 - 50000*1.04^n

Reply 4

arslan

2

ScipioAfricanus

A mortage is taken for £80,000 .It is paid by annaul instalments of £5,000 with the first payment being made at the end of the first year that the mortage was taken out. Interest of 4% is then charged on the out standing debt. Find the the the total time taken to pay off the mortage.

Note: Find an experession for the debt remaining after n years and slove using the fact that if it is paid off, the debt = 0.

thanks for the help in advance. Can you explian how you came to finding the experession.

lot of working out below so it should be clear what is happening:

Debt at the end of 1st year = £80000-5000

Debt at start of 2nd year = (80000-5000)*1.04
= 80000*1.04 - 5000*1.04

Debt at end of 2nd year = (80000*1.04 - 5000*1.04) - 5000

Debt at start of 3rd year = (80000*1.04-5000*1.04-5000)*1.04
= 80000*1.04^2 - 5000*1.04^2 - 5000*1.04
= 80000*1.04^2 - 5000(1.04^2+1.04)

Debt at the end of 3rd year = 80000*1.04^2 - 5000(1.04^2+1.04+1)

so debt at end of nth year =

80000*1.04^n-1 / 5000(1.04^n-1 + 1.04^n-2 +...+ 1.04 + 1)

debt after n years = 0

=> 80000*1.04^n-1 = 5000(1.04^n-1 + 1.04^n-2 +...+ 1.04 +1)
80000*1.04^n-1 = 5000* [1(1.04^n-1)/1.04-1]
80000*1.04^n-1 = [5000*(1.04^n-1)]/0.04
80000*1.04^n-1 = 125000(1.04^n-1)
80000*1.04^n-1 = 125000*1.04^n -125000
80000*1.04^n-1 = 125000*1.04 * 1.04^n-1 - 125000
125000 = 130000 * 1.04^n-1 -80000 * 1.04^n-1
125000 = 50000*1.04^n-1
125000/50000 = 1.04^n-1
2.5 = 1.04^n-1
-----take logs
log 2.5 = log 1.04^n-1
log 2.5 = (n-1) *log 1.04
log 2.5/log 1.04 = n-1
n-1 = 23.3624
therefore n = 24.36

hence total time to pay off debt is 25 years

C2 geometric series

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