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Prize Money

Date: 09/04/97 at 14:27:30
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

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!   
Associated Topics:
High School Basic Algebra
High School Sequences, Series

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