Chapter 9, Problem 5CT

To determine

To calculate: The $n\text{th}$ term of the geometric sequence whose second term is 14 and sixth term is 224.

Answer to Problem 5CT

The $n\text{th}$ term of the arithmetic sequence whose $5\text{th}$ term is 24 and $12\text{th}$ term is 24 is $a_n=\frac{7}{2}{\left(2\right)}^n$ .

Explanation of Solution

Given information:

The geometric sequence whose second term is 14 and sixth term is 224.

Formula used:

The $n\text{th}$ term of the geometric sequence is denoted by $a_n=a{r}^{n-1}$ , where a is the first term, r is the common ratio and n is number of terms.

Calculation:

Consider the geometric sequence whose second term is 14 and sixth term is 224.

Recall that the $n\text{th}$ term of the geometric sequence is denoted by $a_n=a{r}^{n-1}$ , where a is the first term, r is the common ratio and n is number of terms.

Second term of sequence that is $a_2=14$ .

  $\begin{array}{c}{a}_2=14\\ a{r}^{2-1}=14\\ ar=14\end{array}$

Therefore, $ar=14$  ...(1)

Sixth term of sequence that is.

  $\begin{array}{c}{a}_6=224\\ a{r}^{6-1}=224\\ a{r}^5=224\end{array}$

Therefore, $a{r}^5=224$  ...(2)

Solve equation (1) and (2).

From equation (1), value of ais $a=\frac{14}{r}$ . Now substitute the value of a in equation (2),

  $\begin{array}{c}\left(\frac{14}{r}\right)r^5=224\\ r^4=\frac{224}{14}\\ r^4=16\\ r=2\end{array}$

Now substitute the value of r in the equation $a=\frac{14}{r}$ .

  $\begin{array}{c}a=\frac{14}{2}\\ a=7\end{array}$

Last substitute value of a and rin $a_n=a{r}^{n-1}$ ,

  $\begin{array}{c}{a}_n=a{r}^{n-1}\\ =7{\left(2\right)}^{n-1}\\ =\frac{7}{2}{\left(2\right)}^n\end{array}$

Thus, the $n\text{th}$ term of the arithmetic sequence whose $5\text{th}$ term is 24 and $12\text{th}$ term is 24 is $a_n=\frac{7}{2}{\left(2\right)}^n$ .