Geometric sequences

OP

Is there a general rule for which one to use?

Maybe this question is a good example.

A mortgage is taken out for £80000. It is to be paid back by annual instalments of 5000 with the first payment being made at the end of the first year. Interest of 4% is then paid on any outstanding debt. Find the time taken to pay off the mortage.

I have my geometric sequence as

80,000 x 1.04^n - 5000x1.04^n - 5000 x 1.04^n-1 - 5000x 1.04^n-2....

80,000 x 1.04^n = 5000(1.04^n + 1.04^n-1 + 1.04^n-2..)

Am I right in saying that it is in fact n and not n-1, that a = 1.04 and r is also = to 1.04....

I am sooo stuck on it! I cant seem to get the answer of 25 years!

I have no idea what to do and i cant move on until i undersand why!!

Please help

Reply 3

Original post by christinajane

Is there a general rule for which one to use?

Maybe this question is a good example.

A mortgage is taken out for £80000. It is to be paid back by annual instalments of 5000 with the first payment being made at the end of the first year. Interest of 4% is then paid on any outstanding debt. Find the time taken to pay off the mortage.

I have my geometric sequence as

80,000 x 1.04^n - 5000x1.04^n - 5000 x 1.04^n-1 - 5000x 1.04^n-2....

80,000 x 1.04^n = 5000(1.04^n + 1.04^n-1 + 1.04^n-2..)

Am I right in saying that it is in fact n and not n-1, that a = 1.04 and r is also = to 1.04....

I am sooo stuck on it! I cant seem to get the answer of 25 years!

I have no idea what to do and i cant move on until i undersand why!!

Please help

ok sounds like your still here so i'll give a walk through and an answer and method if you really can't figure it out

So find values for

S_n, n, a\ and\ r

the formula for the sum of a geometric series is given by (these will be on formula sheet so no need to worry^-^)

S_n = \dfrac {a\left(r^n -1\right)}{r-1}

or

S_n = \dfrac {a\left(1-r^n\right)}{1-r}

sub in your values of

S_n, n, a\ and\ r

then rearrange until you have

something^n

= a number
then once you've done that take the log of both sides
use the power rule and use calculator to solve

done

(edited 8 years ago)

Reply 4

@Student403 i seem to get the answer 12.6 when i solve for n what did you get?

Reply 5

OP

I got the same but its not right- should be 25....

I know those formulas and have been using them but im obvioulsy putting the wrong values in somewhere....

a = 1.04 r = 1.04 and sn = 80000x1.04^n or something....

(edited 8 years ago)

Reply 6

Original post by christinajane

I got the same but its not right- should be 25....

I know those formulas and have been using them but im obvioulsy putting the wrong values in somewhere....

a = 1.04 r = 1.04 and sn = 80000x1.04^n or something....

if you multiply by 2 you get the right answer :biggrin:

xD but seriously :colonhash:

not sure what's going on here

also shouldn't a be 5000?
and Sn be 80000?

Reply 7

Original post by christinajane

Is there a general rule for which one to use?

Maybe this question is a good example.

A mortgage is taken out for £80000. It is to be paid back by annual instalments of 5000 with the first payment being made at the end of the first year. Interest of 4% is then paid on any outstanding debt. Find the time taken to pay off the mortage.

I have my geometric sequence as

80,000 x 1.04^n - 5000x1.04^n - 5000 x 1.04^n-1 - 5000x 1.04^n-2....

80,000 x 1.04^n = 5000(1.04^n + 1.04^n-1 + 1.04^n-2..)

Am I right in saying that it is in fact n and not n-1, that a = 1.04 and r is also = to 1.04....

I am sooo stuck on it! I cant seem to get the answer of 25 years!

I have no idea what to do and i cant move on until i undersand why!!

Please help

This example is not an easy one. You might want to find a simpler one if it's purely the logs or n/n-1 you want to clarify

But you should try to solve it.

Try writing out the current debt status at the beginning and end of each year. Try to pick out a pattern

(edited 8 years ago)

Reply 8

OP

Its a ridiculous question I agree!

I have the nth pattern at least i think I do...

80000-5000x1.04^n - 5000x1.04^n - 50000 x 1.04^n-1 - 50000x 1.04^n-2

factorised

80000-5000x1.04^n -5000(1.04^n - 1.04^n-1 - 1.04^n-2....) = 0

so then

80000-5000x1.04^n = (1.04^n + 1.04^n-1 + 1.04^n-2...)

Im not sure if this is correct .... and what a should be equal too...

anyway the my onle saving grace is is that Im sure there will be no questions like this on the c2 exam! I hope.

Its more just annoying me that I cant do it....

Reply 9

Careful - not quite, spotted a couple of mistakes. Could you show me what you did before that?

Reply 10

OP

I just tried to write a couple of terms out frst to spot the pattern so...

first year = 80000 - 5000
2nd year = (80000 - 5000) x 1.04
end 2nd year = (800000 - 5000) x 104 - 5000
= 80000 x 1.04 - 5000 x 1.04 - 5000
3rd year = (80000 x 1.04 - 5000 x 1.04 - 5000) x1.04
end of 3rd = (8000 x 1.04^2 - 5000x 1.04^n2 - 5000 x 1.04

If this is what you meant?

oops forgot the 5000 after the = sign in my last post after

80000-5000x1.04^n = 5000(1.04^n + 1.04^n-1 + 1.04^n-2...)

(edited 8 years ago)

Reply 11

Not quite :/ That resembles what it's going to come down to but I suspect you've made a mistake quite a bit earlier. Mind posting a pic of all your working?

Reply 12

Original post by Student403

Not quite :/ That resembles what it's going to come down to but I suspect you've made a mistake quite a bit earlier. Mind posting a pic of all your working?

of course

Spoiler

(edited 8 years ago)

Reply 13

Original post by thefatone

of course

Spoiler

Not that simple. You can't just put values in to the formula

Reply 14

Original post by Student403

Not that simple. You can't just put values in to the formula

oh i see bc the interest on unpaid stuff increases.... right.. i'll do this tomorrow if there's no answer by then, i really wanna do it xD

Reply 15

OP

Original post by Student403

Not quite :/ That resembles what it's going to come down to but I suspect you've made a mistake quite a bit earlier. Mind posting a pic of all your working?

I have so much working out - that when I look back on it it all looks the same - it pretty much follows what I wrote earlier.

I think because I have been trying to do it for so long that my mind can't see anything different to what Ive been doing.

I wrote a few years out just to build a pattern up then tried to put it in a geometric sequence - in the book it gives a hint and says to equal it to zero....

so that why I end up with that equation.....80000-5000x1.04^n = 5000(1.04^n + 1.04^n-1 + 1.04^n-2...) after movinf the RHS to the RHS so the minuses then become pluses....

So - Im not sure you can simply put those numbers in to the Sn formula because of that LHS.

Any pointers would be great - i can't seem to think where IVE gone wrong.

Then I thought instead of setting it to zero - why not make the Sn = to 80000 and then find n.....but my brain won't work. Wish it was more mathematical - I know I am missing something obvious probably :-(

(edited 8 years ago)

Reply 16

It's not bleeding obvious but I suspect you went wrong writing out your year sequences :redface:

Reply 17

OP

Hmmmmm

I can't seem to think where :-( any other clues???

Reply 18

Original post by christinajane

Hmmmmm

I can't seem to think where :-( any other clues???

Okay this might help you get it

DEBT:

Start of the first year is 80000GBP, end of the first year takes off 5000 and multiplies final value by 1.04. end of the second year takes off 5000 and multiplies THAT final value by 1.04.

Now rewrite this mathematically and show me what you get :smile:

Reply 19

OP