Chapter 9, Problem 5CT

To determine

To find: The $n^{th}$ term of the given sequence.

Answer to Problem 5CT

The $n^{th}$ term is $a_n=7\left(4^{n-1}\right)$.

Explanation of Solution

Given:

The second term of the geometric sequence is 28.

The sixth term is 7168.

Calculation:

The $n^{th}$ term for any geometric sequence is,

  $a_n=a{r}^{n-1}$

where, $a$ is the first term of the sequence and $r$ is the common ratio.

Since the second term is 28,

  $\begin{array}{c}{a}_2=a{r}^{2-1}\\ 28=ar\\ a=\frac{28}{r}\end{array}$...........(1)

Since the twelfth term is 11.0,

  $\begin{array}{c}{a}_6=a{r}^{6-1}\\ 7168=a{r}^5\end{array}$

Substitute the value of $a$ from equation (1) in the above equation to determine the value of $r$.

  $\begin{array}{c}7168=\frac{28}{r}{r}^5\\ \frac{7168}{28}=r^4\\ 4^4=r^4\\ r=4\end{array}$

Substitute 4 for $r$ in equation (1) to find the value of $a$.

  $\begin{array}{c}a=\frac{28}{r}\\ =\frac{28}{4}\\ =7\end{array}$

So, the $n^{th}$ term of the sequence is $a_n=7\left(4^{n-1}\right)$.