Chapter 12.2, Problem 5CFU

To determine

To find whether the given terms form a finite geometric sequence.

Answer to Problem 5CFU

No.

Explanation of Solution

Given:

The terms are $3,6,18$ .

Calculation:

Consider $a_1=3,a_2=6,a_3=18$

Now

  $\begin{array}{l}{a}_1=3,a_2=6,a_3=18\\ \frac{a_2}{a_1}=\frac{6}{3}=2\\ \frac{a_3}{a_2}=\frac{18}{6}=3\end{array}$

Therefore $\frac{a_2}{a_1}\ne \frac{a_3}{a_2}$

Hence the sequence is not geometric sequence.

To determine

To find whether the given terms form a finite geometric sequence.

Answer to Problem 5CFU

Yes.

Explanation of Solution

Given:

The terms are $\sqrt{3},3,\sqrt{27}$ .

Calculation:

Consider $a_1=\sqrt{3},a_2=3,a_3=\sqrt{27}$

Now

  $\begin{array}{l}\frac{a_2}{a_1}=\frac{3}{\sqrt{3}}=\sqrt{3}\\ \frac{a_3}{a_2}=\frac{\sqrt{27}}{3}=\sqrt{3}\end{array}$

Therefore $\frac{a_2}{a_1}=\frac{a_3}{a_2}$

Hence the sequence is geometric sequence.

To determine

To find whether the given terms form a finite geometric sequence.

Answer to Problem 5CFU

Yes.

Explanation of Solution

Given:

The terms are $x^{-2},x^{-1},1$ .

Calculation:

Consider $a_1=x^{-2},a_2=x^{-1},a_3=1$

Now

  $\begin{array}{l}\frac{a_2}{a_1}=\frac{x^{-1}}{x^{-2}}=x\\ \frac{a_3}{a_2}=\frac{1}{x^{-1}}=x\end{array}$

Therefore $\frac{a_2}{a_1}=\frac{a_3}{a_2}$

Hence the sequence is geometric sequence.