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I could use some help with the following:

Jacques is saving for a new car which will cost 29000 dollars. He saves by putting 400 dollars a month into a savings account which gives 0.1% interest per month. After how many months will he be able to buy his car? (assume it does not increase in price)

Soo at the end of the first month, what he has is 400*1.1 + 400; and at the end of the 2nd month this is 840*1.1 + 400 . But I can't really find the way to express this as geometric series. If n is the # of months, r the common ratio it'll be like [400[(r)^n-1]]/(r-1)

..but how exactly? -excuse my amateurishness, its been a while- thanks in advance!

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up vote 3 down vote accepted

$0.1$% is $0.001$, so the monthly multiplier is $1.001$, not $1.1$.

If $n$ is the number of months, $a$ is the monthly deposit ($\$400$ in your problem), and $r$ is the multiplier ($1.001$ in your problem), the first deposit grows to $ar^{n-1}$, since it’s compounded $n-1$ times, the second to $ar^{n-2}$, and so on, down to the $n$-th (last) deposit, which ‘grows’ to $a$, since it’s not compounded at all. You therefore have $$a+ar+ar^2+\ldots+ar^{n-2}+ar^{n-1}\;,$$ which is indeed a geometric series. You almost remembered the formula for its sum:


In your problem that comes to $$\frac{400(1.001^n-1)}{1.001-1}=\frac{400(1.001^n-1)}{0.001}=400,000(1.001^n-1)\;.$$

If $C$ is the cost of the car, you want to solve $400,000(1.001^n-1)=C$ for $n$. After some basic algebra you have $$1.001^n=\frac{C}{400,000}+1\;;$$ if you now take logarithms on both sides you have $$n\log 1.001=\log\left(\frac{C}{400,000}+1\right)\;,$$ which is easily solved for $n$.

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cheers, how careless of me to write 0.1% as 1/10. I was serious when I said its been a while. – Deniz May 14 '12 at 7:32

Write the amount at the end of the second month as $((400)(1.1)+400)(1.1)+400$, which is $(400)(1.1)^2+(400)(1.1)+400$; at the end of the third month, it will be $$(400)(1.1)^3+(400)(1.1)^2+(400)(1.1)+400$$ and so on. Do you see the geometric series?

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