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Josh wanted to buy a bike but didn't have enough money
Josh wanted to buy a bike but didn't have enough money. Sam slick said "I can fix that. Each time you jump that fence, I'll double your money. There's one small thing though. Each time I pay you, you must give me $56 back for the privilege of jumping." Josh agreed, jumped the fence, received his payment from Sam slick, and paid Sam $56. Repeating the routine 2 more times, josh was distressed to find that, on the last jump after Sam had made his payment to josh, josh had only $56 with which to pay Sam and so had nothing left. Sam of course , went merrily on his way leaving josh wishing that he had known a little more about mathematics.
(A) josh had $ ____ before he made his deal with Sam.
(B) if josh jumps the fence 4 times before running out of money, then he has $_____ before he made his deal with Sam
Expert Solution
Answer:
a) $49
b) $52.50
Step-by-step explanation
a) The amount that Josh has left after each jump is an = 2an-1 - 56. This has the form an = ran-1 + c, which can be rewritten in the form
an = A*rn + ∑c*rn-k. This has the solution an = Arn + c[(rn - 1)/(r - 1)], when r ≠ 1. In this case, r = 2, so that solution applies. The value of A is the initial amount that Josh started with. The value of c is -56. It is also known that a3 = 0. Substituting a3 = 0, r = 2, n = 3, and c = -56 and solving for A gives:
an = Arn + c[(rn - 1)/(r - 1)]
0 = A(23) + (-56)[ (23 - 1) / (2 - 1)]
0 = 8A + (-56)(7)
8A = 56(7)
A = 56(7)/8
A = 49
Thus, Josh started with $49.
b) In this case, the same equation applies: an = Arn + c[(rn - 1)/(r - 1), except that n = 4. Substituting n = 4, r = 2, and c = -56 again, and knowing that a4 = 0, solving for A gives:
0 = A(24) + (-56)[ (24 - 1) / (2 - 1)]
0 = 16A + (-56)(15)
16A = (56)(15)
A = 56(15)/16
A = 52.50
Thus, in this case, Josh started with $52.50.
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