The first date on which the received amount is less than 1 cent and the total amount received up to that day, if a rich man gives $ 1000 on September 1 , and thereafter he will give 9 10 of what he gave the previous day.
The first date on which the received amount is less than 1 cent and the total amount received up to that day, if a rich man gives $ 1000 on September 1 , and thereafter he will give 9 10 of what he gave the previous day.
Solution Summary: The author calculates the amount received after n number of days.
To calculate: The first date on which the received amount is less than 1 cent and the total amount received up to that day, if a rich man gives $1000 on September 1, and thereafter he will give 910 of what he gave the previous day.
Expert Solution & Answer
Answer to Problem 101AYU
Solution:
On the 111th day or on December20, the amount will be less than $0.01
Up to 111th day, the total amount will become $9999.92
Explanation of Solution
Given information:
A rich man gives $1000 on September 1, and thereafter he will give 910 of what he gave the previous day.
Formula used:
Let {an} be a geometric sequence with first term a1 and common ratio r, where r≠0,1. The sum Sn of the first n terms of {an} is
Sn=a1+a1r+a1r2+⋅⋅⋅+a1rn−1=∑k=1na1rk−1=a1⋅1−rn1−r
Calculation:
Assume the received amount is less than 1 cent after n number of days from September 1
The amount received after n number of days can be calculated by an=arn−1
Here, a=1000 and r=910=0.9
⇒1000(0.9)n−1<0.01
⇒(0.9)n−1<0.011000
⇒(0.9)n−1<0.00001
By taking ln on both sides,
⇒(n−1)ln(0.9)<ln(0.00001)
⇒(n−1)(−0.105361)<(−11.512926)
⇒n−1>−11.512926−0.105361
⇒n−1≈109.27
⇒n=109.27+1
⇒n=110.27
Thus, on the 111th day or December 20, the amount will be less than $0.01.
Now, by substituting n=111 in the sum of first n terms of geometric series formula Sn=a11−rn1−r,
⇒S111=1000⋅1−(0.9)1111−(0.9)
⇒S111=1000⋅1−(0.9)1110.1
⇒S111=9999.92
Thus, up to 111th day, the total amount will become $9999.92
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