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### Prize Money

```
Date: 09/04/97 at 14:27:30
From: DOMINIQUE VICKERS
Subject: Sequential order

We have prize money totaling \$6,000. If first place wins \$1,000 and
twentieth place wins \$100, how much money do second through
nineteenth place win? It seems to be a geometric sequence, but we
cannot figure it out using the limits.
```

```
Date: 09/10/97 at 13:50:03
From: Doctor Rob
Subject: Re: Sequential order

If it is a geometric progression, then it has the form
a, a*r, a*r^2, ... where the n-th term is a*r^(n-1).
The first and 20th terms will suffice to figure out a and r.
The sum is then a*(1-r^n)/(1-r).  In this case,
a = \$1000, r = 10^(-1/19) = 0.88586679, and the sum is \$7985.52.

This seems to show that the prize money does not obey a geometric
progression.

If it is an arithmetic progression, then it has the form
a, a+d, a+2*d, ..., where the n-th term is a + (n-1)*d.
The sum is n*a+d*n*(n-1)/2. In this case, a = \$1000, d = -\$900/19,
and the sum is 20*\$1000 - (\$900/19)*20*19/2 = \$20000 - \$9000 = \$11000.
This seems to show that the prize money does not obey an arithmetic
progression, either.

If it has the form a + b*r^n, then a+b = 1000, a+b*r^19 = 100, and
the sum 20*a + b*(1-r^20)/(1-r) = 6000, you will have three equations
in three unknowns, which you should be able to solve.

a = 1000 - b,
r = [(b-900)/b]^(1/19),
20000 - 20*b + b*(1-r*[b-900]/b)/(1-r) = 6000
20000 - 20*b + (b-r*b+900*r)/(1-r) = 6000
14000 + 900*r/(1-r) = 19*b
b = (14000-13100*r)/(1-r)
b = 13100 + 900/(1-r)
1 - r = 900/(b-13100)
r = 1 - 900/(b-13100)
r = (b-14000)/(b-13100)

There are several values of b that work:  b = 14007.2257860276,
14891.0671719881, 15225.4844265359, 17199.53307229. Using these
and the above equations, you can figure out r and a.

If you assume another 3-parameter formula for the prize money, you
can probably make that work, too.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Sequences, Series

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