Chapter 11.3, Problem 36SR

To determine

To find: The first five terms of geometric sequence.

Answer to Problem 36SR

The first five terms are $21$ , $14$ , $\frac{28}{3}$ , $\frac{56}{9}$ , and $\frac{112}{27}$ .

Explanation of Solution

Given:

It is given that $a_1=21,r=\frac{2}{3}$ .

Calculation:

The formula for geometric sequence is given by $a_n=a_1{r}^{n-1}$ .

Substitute 21 for $a_1$ , $\frac{2}{3}$ for $r$ and $2$ for $n$ in the formula $a_n=a_1{r}^{n-1}$ to find the second term.

  $\begin{array}{c}{a}_2=21{\left(\frac{2}{3}\right)}^{2-1}\\ =14\end{array}$

Substitute 21 for $a_1$ , $\frac{2}{3}$ for $r$ and 3 for $n$ in the formula $a_n=a_1{r}^{n-1}$ to find the third term.

  $\begin{array}{c}{a}_3=21{\left(\frac{2}{3}\right)}^{3-1}\\ =\frac{28}{3}\end{array}$

Substitute 21 for $a_1$ , $\frac{2}{3}$ for $r$ and 4 for $n$ in the formula $a_n=a_1{r}^{n-1}$ to find the fourth term.

  $\begin{array}{c}{a}_4=21{\left(\frac{2}{3}\right)}^{4-1}\\ =\frac{56}{9}\end{array}$

Substitute 21 for $a_1$ , $\frac{2}{3}$ for $r$ and 5 for $n$ in the formula $a_n=a_1{r}^{n-1}$ to find the fifth term.

  $\begin{array}{c}{a}_5=21{\left(\frac{2}{3}\right)}^{5-1}\\ =\frac{112}{27}\end{array}$

Therefore, the first five terms are $21$ , $14$ , $\frac{28}{3}$ , $\frac{56}{9}$ , and $\frac{112}{27}$ .