Chapter 7.3, Problem 7MRPS

Solution

a.

To determine : if the sequence is arithmetic, geometric or neither if a scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

The sequence $66,33,16.5,8.25,\dots$ is a geometric series.

Given information :

A scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

Formula used :

The common ratio of a geometric series is calculated as $r=\frac{a_n}{a_{n-1}}$ .

Calculation :

Consider the sequence $66,33,16.5,8.25,\dots$ ,

Calculate the value of $r=\frac{a_n}{a_{n-1}}$ .

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{33}{66}\\ =0.5\end{array}$

  $\begin{array}{c}r=\frac{a_3}{a_2}\\ =\frac{16.5}{33}\\ =0.5\end{array}$

  $\begin{array}{c}r=\frac{a_4}{a_3}\\ =\frac{8.25}{16.5}\\ =0.5\end{array}$

As the common ratio is same, the series is a geometric series.

Thus, the sequence $66,33,16.5,8.25,\dots$ is a geometric series.

b.

To calculate : the rule for $n^{th}$ term of the sequence if a scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

The rule for $n^{th}$ term of the sequence $66,33,16.5,8.25,\dots$ is $a_n=a_1{0.5}^{n-1}$ .

Given information :

A scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

Formula used :

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

Calculation :

Consider the sequence $66,33,16.5,8.25,\dots$ ,

From part (a), the value of $r$ is $0.5$ .

Put the value $r=0.5$ into the formula $a_n=a_1{r}^{n-1}$ .

  $a_n=a_1{0.5}^{n-1}$

Thus, the rule for $n^{th}$ term of the sequence $66,33,16.5,8.25,\dots$ is $a_n=a_1{0.5}^{n-1}$ .

c.

To graph: the sequence if a scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

Given information:

The given sequence is $66,33,16.5,8.25,\dots$ .

Graph:

Consider the given sequence:

  $66,33,16.5,8.25,\dots$

The terms represent the radioactive decay every two hours. So, the first hour can be considered as point 1.

These terms can be written as: $\left(1,66\right),\left(2,33\right),\left(3,16.5\right),\left(4,8.5\right)$

Now to draw the graph by plotting these points on the graph:

  

Interpretation: It can be seen that as $n$ increases, the decay also increases, but the value approaches to $0$ as $n\to \infty$ as there will be complete decay.

d.

To calculate : the time after which the scientist measures Platinum-197 that is less than 1 gram if a scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

The time after which the scientist measures Platinum-197 that is less than 1 gram is at most 7 hours.

Given information :

A scientist is studying the radioactive decay of Platinum-197. The scientist starts with 66-gram sample of Platinum-197 and then measures the amount remaining every two hours. The amount recorded are $66,33,16.5,8.25,\dots$ .

Formula used :

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

Calculation :

From part (b), the rule comes out to be $a_n=a_1{0.5}^{n-1}$ .

It is needed to find a number such that $a_n<1$ .

  $\begin{array}{c}{a}_n<1\\ a_1{0.5}^{n-1}<1\\ 66{\left(0.5\right)}^{n-1}<1\\ {\left(0.5\right)}^{n-1}<\frac{1}{66}\end{array}$

Take ${\log }_{0.5}$ on both sides.

  $\begin{array}{c}{\log }_{0.5}{\left(0.5\right)}^{n-1}<\log \frac{1}{66}\\ n-1{\log }_{0.5}0.5<{\log }_{0.5}\frac{1}{66}\\ n-1<6.044\\ n<7.044\end{array}$

The time required is at most 7 hours.

Thus, the time after which the scientist measures Platinum-197 that is less than 1 gram is at most 7 hours.