Chapter 9.8, Problem 3CYP

To determine

The equation for nth term and the tenth term of the sequence

Answer to Problem 3CYP

The equation for nth term is $a_n=96\times {\left(\frac{1}{2}\right)}^{n-1}$

The tenth term of the sequence is $a_{10}=\frac{96}{512}$

Explanation of Solution

Given information:

Sequence $96,48,24,12,\mathrm{...}$

Concept Used:

The ratio is given by the formula

  $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$

Calculation:

In order to find the sequence $A_n$ and the ratio between two number r and simplify further as shown below,

  $\begin{array}{c}\text{Common}\text{diffrence}\\ r=\frac{a_2}{a_1}\\ r=\frac{48}{96}\\ =\frac{1}{2}\end{array}$

The ratio of consecutive numbers is $\frac{1}{2}$

Thus, the equation for nth term is given below

  $\begin{array}{l}{a}_1=96\\ r=\frac{1}{2}\\ a_n=a_1{\left(r\right)}^{n-1}\\ a_n=96\times {\left(\frac{1}{2}\right)}^{n-1}\end{array}$

The equation for nth term is $a_n=96\times {\left(\frac{1}{2}\right)}^{n-1}$

The tenth term of the sequence is below

  $\begin{array}{c}{a}_1=96\\ r=\frac{1}{2}\\ a_n=a_1{\left(r\right)}^{n-1}\\ a_n=96\times {\left(\frac{1}{2}\right)}^{n-1}\\ n=10\\ a_{10}=96\times {\left(\frac{1}{2}\right)}^{10-1}\\ =96\times {\left(\frac{1}{2}\right)}^9\\ =96\times \frac{1}{512}\\ a_{10}=\frac{96}{512}\end{array}$

Thus, the tenth term of the sequence is $a_{10}=\frac{96}{512}$