Chapter 11.3, Problem 4ST

(a)

To determine

To calculate:

The formula for $n^{th}$ term for the geometric sequence $48,-24,12,-6,\mathrm{...},$

Answer to Problem 4ST

The formula for $n^{th}$ term for the geometric sequence $48,-24,12,-6,\mathrm{...},$ is $t_n=\left(48\right){\left(-\frac{1}{2}\right)}^{n-1}$ .

Explanation of Solution

Given information:

  $48,-24,12,-6,\mathrm{...},$

Formula used:

Common ratio: $r=\frac{t_{n-1}}{t_n}$

Geometric progression: $t_n=t_1{r}^{n-1}$

Calculation:

Consider the sequence as:

  $\begin{array}{l}{a}_1=48\\ a_2=-24\\ a_3=12\\ a_4=-6\end{array}$

Now, the common ratio is calculated as:

  $\begin{array}{l}r=\frac{a_2}{a_1}=\frac{-24}{48}=-\frac{1}{2}\\ r=\frac{a_3}{a_2}=\frac{12}{-24}=-\frac{1}{2}\\ r=\frac{a_4}{a_3}=\frac{-6}{12}=-\frac{1}{2}\end{array}$

Now, the $n^{th}$ term is calculated as:

  $\begin{array}{l}{t}_n=t_1{r}^{n-1}\\ t_n=\left(48\right){\left(-\frac{1}{2}\right)}^{n-1}\end{array}$

(b)

To determine

To calculate:

The value of $t_{10}$ term for the geometric sequence $48,-24,12,-6,\mathrm{...},$

Answer to Problem 4ST

The value of $t_{19}$ term for the geometric sequence $48,-24,12,-6,\mathrm{...},$ is $-0.09375$ .

Explanation of Solution

Given information:

  $48,-24,12,-6,\mathrm{...},$

Calculation:

The $n^{th}$ term formula from sub-part (a):

  $t_n=\left(48\right){\left(-\frac{1}{2}\right)}^{n-1}$

Here, $n=10$

  $\begin{array}{l}{t}_{10}=\left(48\right){\left(-\frac{1}{2}\right)}^{10-1}\\ t_{10}=\left(48\right){\left(-\frac{1}{2}\right)}^9\\ t_{10}=-0.09375\end{array}$