Chapter 12.2, Problem 42E

a.

To determine

To calculate:

Common ratio and first term of the sequence.

Answer to Problem 42E

Common ratio for given series is $3^{1/3}$ and first term is $\frac{4}{{\left({\left(3\right)}^{1/3}\right)}^3}$ .

Explanation of Solution

Given information:

  $a_4=4,a_7=12$

Calculation:

It is given that, the geometric sequence has $a_4=4,a_7=12$ .

Apply the formula for the $n^{th}$ term of geometric series.

  $a_n=a_1{r}^{n-1}$

The forth term of the geometric series is,

  $\begin{array}{l}{a}_4=a_1{r}^{4-1}\\ 4=a_1{r}^3\end{array}$

The seventh term of the geometric series is,

  $\begin{array}{l}{a}_7=a_1{r}^{7-1}\\ 12=a_1{r}^6\end{array}$

From above two equation,

  $\begin{array}{l}\frac{12}{4}=\frac{a_1{r}^6}{a_1{r}^3}\\ 3=r^3\\ r=3^{1/3}\end{array}$

From forth term, first term is,

  $\begin{array}{l}{a}_1=\frac{4}{r_3}\\ a_1=\frac{4}{{\left({\left(3\right)}^{1/3}\right)}^3}\end{array}$

Thus, Common ratio for given series is $3^{1/3}$ and first term is $\frac{4}{{\left({\left(3\right)}^{1/3}\right)}^3}$ .

b.

To determine

To calculate:

Twenty eighth term of the sequence.

Answer to Problem 42E

Twenty eighth term of the sequence is $26244$ .

Explanation of Solution

Given information:

  $a_4=4,a_7=12$

Calculation:

It is given that, the geometric sequence has $a_4=4,a_7=12$ .

Apply the formula for the $n^{th}$ term of geometric series.

  $a_n=a_1{r}^{n-1}$

The twenty eighth term of the geometric series is,

  $\begin{array}{l}{a}_{28}=a_1{r}^{28-1}\\ 4=\left(\frac{4}{3}\right){\left({\left(3\right)}^{1/3}\right)}^{27}\\ =26244\end{array}$

Thus, Twenty eighth term of the sequence is $26244$ .