kryton1212
Mr. Chan will deposit $25000 in a bank account at the beginning of each year from 2007 to 2012. If the interest is compounded yearly and the interest rate is 4% p.a., find the total amount that he can get at the beginning of 2013. (corr. to 3 sig. fig.)
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kryton1212
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@hartnn
ganeshie8
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looks like annuity setup
ganeshie8
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P = 25,000
r = .04
n = 6
ganeshie8
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\(A = P[ (1+r)^n + (1+r)^{n-1} + .... (1+r)]\)
= \(P[\frac{1-(1+r)^n }{ 1-(1+r)}]\)
ganeshie8
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plug and chug..
kryton1212
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ahhh....wait
ganeshie8
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ha take ur time not an easy problem :)
primeralph
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|dw:1372927372461:dw|
kryton1212
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errr...my calculator shows that it's 165824.3866
primeralph
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@ganeshie8 Why the long stuff?
ganeshie8
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every year he is adding 25 grand
kryton1212
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@primeralph this's a question about sequence
primeralph
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Well, both ways work.
Mine is just shorter.
kryton1212
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@primeralph but your equation does match the answer, neither @ganeshie8 ..
primeralph
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Uh hold on.
ganeshie8
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prime i used ur formula and took a sum to arrive at the final formula
primeralph
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@kryton1212 What's the answer?
kryton1212
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172000 after corr. to 3 sig fig
kryton1212
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uhh?
kryton1212
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i mean the correct answer..
primeralph
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It is not possible to get that much money after just 6 years.
ganeshie8
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im getting 165,824 @kryton1212
prime read the q again
kryton1212
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i don't know how to do...
@hartnn show yourself-_-
ganeshie8
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which formula wont work ?
primeralph
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ohhh..........beginning of each year.
Hold on then.
kryton1212
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i also think that @ganeshie8 's formula is correct...maybe i should ask my teacher about this question.
Let's move on to the next question that you both will be tagged by me a moment later
kryton1212
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...
primeralph
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Hold on.
ganeshie8
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okie
kryton1212
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maybe @primeralph goes on
primeralph
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Uh well, tag away if you're satisfied.
kryton1212
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uh..
ganeshie8
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move on to new problem, im pretty sure what we worked is correct. if you do manually 6 times also, you would get the same answer.
kryton1212
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at the beginning of 2013..................
kryton1212
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@primeralph do you have any other ideas?
primeralph
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To check, you can do each step. If you get the same thing, let it go.
kryton1212
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i am lazy XD...i will ask my teacher about this question tonight.
thank you both
ganeshie8
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np :)
hartnn
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i am getting
172457.362022
kryton1212
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ohh, how?
ganeshie8
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3 sig figs, thats the right answer also... how ?
hartnn
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i calculated for each year...6 calculations :\
hartnn
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2008---25k(1+0.04)=26k
2009---(26k+25k)(1+0.04)=53040
2010---(53040+25k)(1+0.04)=81161.1
2011---(81161.1+25k)(1+0.04)=110408.064
2012---(110408.064+25k)(1+0.04)=140824.38
2013---(140824.38+25k)(1+0.04)=172457....
kryton1212
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wow
ganeshie8
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oh every year rounding off to 3 sigfigs is it
kryton1212
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that means we use the wrong method @ganeshie8 .....
ganeshie8
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may be
kryton1212
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@hartnn million thanks. but how to present it in a shorter form...?
hartnn
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thats what i am thinking...
if we have it for just 2 years,
\(P(1+r)+ [P(1+r)+P](1+r) = P [(1+r)+(1+r)(2+r)]\)
so, maybe for 6 years, the formula will become
\(P (1+r)[1+(2+r)[1+(3+r)[1+(4+r)[1+(5+r)[1+(6+r)]]]]]\)
maybe there's a simplification involved, maybe...
kryton1212
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wow, i think this's more complicated..
hartnn
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hmm.....maybe they wanted us to calculate each year's separately...because i don't see any explicit formula for this...
kryton1212
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uhh, anyway, thank you @hartnn
hartnn
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hmm....welcome ^_^