Question

Julia saved 1 cent ($0.01) on the first day of the month, 2cents ($0.02) on the second day, 4 cents ($0.04) on thethird day and double the previous day's amount on eachsuccessive day Julia's savings for the first five days areshown in the table belowJulia's SavingsDay 1: $0.01Day 2: $0.02Day 3: $0.04Day 4: $0.08Day 5: $0.16on what day will Julia first save an amount greater than$1.00?

Answer 1

The sequence of Julia's savings are a geometric progression, which is a sequence that each term became greter by a constant factor. In this case, que question says that the amount saved doubles each day, starting from $0.01.

If we call a = 0.01 the first term and r = 2 the factor, we can obtain the sum of the amount saved using the formula of the sum of a geometric progression series, which is:

$S_n=\frac{a(r^n-1)}{r-1}$

Where S is the sum and n is the number of term. We know a = 0.01, r = 2 and we want to know in which day (n) we get S = 1.00. Thus:

$\begin{gathered} 1.00=\frac{0.01(2^n-1)}{2-1}=0.01(2^n-1) \\ 2^n-1=\frac{1.00}{0.01}=100 \\ 2^n=100+1=101 \end{gathered}$

If we see the values of 2^n, we get, for n -> 1,2,3,4,5,6,7...:

2^n -> 2,4,8,16,32,64,128

Or, we can say:

$2^7=128>101$

This means that for n = 7, S will be greater than $1.00. We can test that by putting into the formula:

$S_n=\frac{a(r^n-1)}{r-1}=\frac{0.01(2^7-1)}{2-1}=0.01(128-1)=0.01\cdot127=1.27>1.00$

So, we can say that on the 7th day, Julia will first save an amount greater than $1.00.