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/
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