Chapter 12.3, Problem 9AYU

To determine

To find: Prove the sequence geometric and find common ratio along with first four terms.

Answer to Problem 9AYU

The sequence is geometric with common ratio $r=\frac{1}{2}$ and first four terms are . $\frac{-3}{2},-\frac{3}{4},-\frac{-3}{8}$ and $\frac{-3}{16}$ .

Explanation of Solution

Given information:

The given relation is.

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

Calculation:

Consider the given relation.

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

Calculate first four term of the given sequence.

  $\begin{array}{c}{a}_1=-3\times {\left(\frac{1}{2}\right)}^1\\ =\frac{-3}{2}\\ a_2=-3{\left(\frac{1}{2}\right)}^2\\ =\frac{-3}{4}\end{array}$ .

  $\begin{array}{c}{a}_3=-3\times {\left(\frac{1}{2}\right)}^3\\ =\frac{-3}{8}\\ a_4=-3{\left(\frac{1}{2}\right)}^4\\ =\frac{-3}{16}\end{array}$

For the sequence to be geometric this relation must be true.

  $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$

Here $r$ is common ratio.

Put the values and simplify.

  $\begin{array}{c}r=\frac{\frac{-3}{4}}{\frac{-3}{2}}=\frac{\frac{-3}{8}}{\frac{-3}{4}}\\ r=\frac{1}{2}=\frac{1}{2}\end{array}$ .

Therefore, the sequence is geometric with common ratio $r=\frac{1}{2}$ and first four terms are . $\frac{-3}{2},-\frac{3}{4},-\frac{-3}{8}$ and $\frac{-3}{16}$ .