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# Geometric sequences

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1. in geometric sequences , when a log questions comes linking to geometric questions , how do you know what to use , n or n-1??
2. Moved to maths, can you give us a more concrete example?
3. 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!!

4. (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!!

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

or

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

done
5. Student403 i seem to get the answer 12.6 when i solve for n what did you get?
6. 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....
7. (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 xD but seriously not sure what's going on here

also shouldn't a be 5000?
and Sn be 80000?
8. (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!!

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
9. 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....
10. Careful - not quite, spotted a couple of mistakes. Could you show me what you did before that?
11. 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...)
12. 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?
13. (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:
Show

and i got 12.6
if i multiply by 2 i get 25 this is not right though like you said
14. (Original post by thefatone)
of course
Spoiler:
Show
Not that simple. You can't just put values in to the formula
15. (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
16. (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 :-(
17. It's not bleeding obvious but I suspect you went wrong writing out your year sequences
18. Hmmmmm

I can't seem to think where :-( any other clues???
19. (Original post by christinajane)
Hmmmmm

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

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
20. Ok let me try :-)

first year- 80,000-5000
= 75000

end first = 75000 x 1.04

start of 2nd = 78000 - 5000
= 73000

end of 2nd = 73000 x 1.04
= 75920

=75000 x 1.04^n - 5000 x 1.04 - 5000 x 104
=75000 x 1.04^n - 5000(1.04+1.04^2+1.04^3+1.04^n)

oooo dont laugh if im wildly off track....

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