Re: geometric progressions

On May 8, 7:26 pm, Sebastian <promil...@xxxxxxxx> wrote:
Judging by the problem statement, I think this should be: 1000*1.12 in
the first cell (end of 1990), then after that,
(1+0.12)*(cell above + 1000). Each cell is cash on hand at year's end.
This accumulates to $19,654.58 by the end of 1999.

I wish the OP would get back to us, and tell us what his answer was
and what was the book's answer.

R.G. Vickson

You are right. I didn't pay attention to any "+-1 matter" as I was
only interested in finding a general solution.
In my example above I stopped the iteration one year early. R.G.
Vicksons formula is otherwise equivalent to mine (except he iterates
over the "ends" while I iterate over the "beginnings" of each year).
Anyway the correct answer would be $19,654.58

Concerning the amount in the year 19999 I can only say that my
spread*** software quit already in the year 8187 at an amount of
1,7213E+308...

Ok, here is the solution as given a(r^n-1)/(r-1)

1000*1.12(1.12^10-1)/
(1.12-1) = $19,654.5832

Thanks to all concerned with their help.

Watch this space...... Probably be back with some more problems over
the year!!!

Martin

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