Question

Show that each sequence is geometric. Then find the common ratio and write out the first four terms$\left\{a_n\right\}=\left\{-3{\left(\frac{1}{2}\right)}^n\right\}$

Short answer

The common ratio is$\frac{1}{2}$ and the first four terms are$\frac{-3}{2},\frac{-3}{4},\frac{-3}{8},\frac{-3}{16}$.

Step-by-step solution

Step 1. Given information.

We have given sequence.

$\left\{a_n\right\}=\left\{-3{\left(\frac{1}{2}\right)}^n\right\}$

The $n$th term and $(n-1)$th term are

role="math" localid="1646849849630" $$\begin{gathered} a_n=-3{\left(\frac{1}{2}\right)}^n \\ {a}_{n-1}=-3{\left(\frac{1}{2}\right)}^{n-1} \end{gathered}$$

Step 2. To find the common ratio. 

We have,

$$\begin{gathered} r=\frac{a_n}{a_{n-1}} \\ r=\frac{-3{\left({\displaystyle \frac{1}{2}}\right)}^n}{-3{\left({\displaystyle \frac{1}{2}}\right)}^{n-1}} \\ r={\left(\frac{1}{2}\right)}^{n-n+1} \\ r=\frac{1}{2} \end{gathered}$$

Since the ratio of the successive term is the non zero constant$\frac{1}{2}$therefore$\left(s_n\right)$ is a geometric sequence with common ratio is$\frac{1}{2}$.

Step 3. To find the terms.  

We have,

$$\begin{gathered} a_1=-3{\left(\frac{1}{2}\right)}^1 \\ {a}_1=\frac{-3}{2} \\ {a}_2=-3{\left(\frac{1}{2}\right)}^2 \\ {a}_2=\frac{-3}{4} \\ {a}_3=-3{\left(\frac{1}{2}\right)}^3 \\ {a}_3=\frac{-3}{8} \\ {a}_4=-3{\left(\frac{1}{2}\right)}^4 \\ {a}_4=\frac{-3}{16} \end{gathered}$$

Thus, the first four terms of the sequence are$\frac{-3}{2},\frac{-3}{4},\frac{-3}{8},\frac{-3}{16}$.