Question

Show that each sequence is geometric. Then find the common ratio and write out the first four terms$\left(f_n\right)=\left\{3^{2n}\right\}$

Short answer

The common ratio is$9$ and the first four terms are$9,81,729,6561$.

Step-by-step solution

Step 1. Given information. 

We have given sequence $\left(f_n\right)=\left(3^{2n}\right)$

The $nth$and $(n-1)th$terms are

$$\begin{gathered} f_n=3^{2n} \\ {f}_{n-1}=3^{2(n-1)} \end{gathered}$$

Step 2. To find the common ratio.    

We have,

$$\begin{gathered} r=\frac{f_n}{f_{n-1}} \\ r=\frac{3^{2n}}{3^{2n-2}} \\ r=3^{2n-2n+2} \\ r=9 \end{gathered}$$

Since the ratio of the successive term is the non zero constant$9$therefore,$\left(f_n\right)$ is a geometric sequence with common ratio is$9$.

Step 3. To find the terms.   

We have,

$$\begin{gathered} f_1=3^{2\left(1\right)} \\ {f}_1=9 \\ {f}_2=3^{2\left(2\right)} \\ {f}_2=81 \\ {f}_3=3^{2\left(3\right)} \\ {f}_3=729 \\ {f}_4=3^{2\left(4\right)} \\ {f}_4=6561 \end{gathered}$$

Thus, the first four terms are$9,81,729,6561$.