Question 213128
$1,500 is deposited every year in an account yielding 6% interest compounded annually. How much money will have been saved after 10 years?
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The formula for the amount,$A an initial investment of 
$P will grow to in t years is given by:

A = P(1+r)^t

The $1500 deposited at the beginning of the 10th year 
will have grown to $1500(1+r)^1 in the 1 year it will 
have been on deposit at the end of the 10th year.

The $1500 deposited at the beginning of the 9th year 
will have grown to $1500(1+r)^2 in the 2 years it will 
have been on deposit at the end of the 10th year.

The $1500 deposited at the beginning of the 8th year 
will have grown to $1500(1+r)^3 in the 3 years it will 
have been on deposit at the end of the 10th year.

...

The $1500 deposited at the beginning of the 2nd year 
will have grown to $1500(1+r)^9 in the 9 years it will 
have been on deposit at the end of the 10th year.

The $1500 deposited at the beginning of the 1st year 
will have grown to $1500(1+r)^10 in the 10 year it will 
have been on deposit at the end of the 10th year.

So the sum is this series:

$1500(1+r) + $1500(1+r)^2 + $1500(1+r)^3 + ... + $1500(1+r)^10

and since r = 6% = .06

$1500(1.06) + $1500(1.06)^2 + $1500(1.06)^3 + ... + $1500(1.06)^10

The common ratio is 1.06, so we use the formula for the sum
of the first n terms of a geometric series:

{{{S[n]=a[1](1-r^n)/(1-r)}}}

Substituting {{{a[1] = "$1500(1.06)"="$1590"}}}, {{{r=1.06}}} and {{{n=10}}}

{{{S[10]="$1590"*(1-1.06^10)/(1-1.06) = "$20957.46")}}} 

Edwin</pre>