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looks like annuity setup

P = 25,000
r = .04
n = 6

\(A = P[ (1+r)^n + (1+r)^{n-1} + .... (1+r)]\)
= \(P[\frac{1-(1+r)^n }{ 1-(1+r)}]\)

plug and chug..

ahhh....wait

ha take ur time not an easy problem :)

|dw:1372927372461:dw|

errr...my calculator shows that it's 165824.3866

@ganeshie8 Why the long stuff?

every year he is adding 25 grand

@primeralph this's a question about sequence

Well, both ways work.
Mine is just shorter.

@primeralph but your equation does match the answer, neither @ganeshie8 ..

Uh hold on.

prime i used ur formula and took a sum to arrive at the final formula

@kryton1212 What's the answer?

172000 after corr. to 3 sig fig

uhh?

i mean the correct answer..

It is not possible to get that much money after just 6 years.

http://www.wolframalpha.com/input/?i=25000%281-%281%2B.04%29%5E6%29%2F%281-%281%2B.04%29%29

im getting 165,824 @kryton1212
prime read the q again

which formula wont work ?

ohhh..........beginning of each year.
Hold on then.

...

Hold on.

okie

maybe @primeralph goes on

Uh well, tag away if you're satisfied.

uh..

at the beginning of 2013..................

@primeralph do you have any other ideas?

To check, you can do each step. If you get the same thing, let it go.

i am lazy XD...i will ask my teacher about this question tonight.
thank you both

np :)

i am getting
172457.362022

ohh, how?

3 sig figs, thats the right answer also... how ?

i calculated for each year...6 calculations :\

wow

oh every year rounding off to 3 sigfigs is it

that means we use the wrong method @ganeshie8 .....

may be

wow, i think this's more complicated..

hmm....welcome ^_^