Question

How do you determine whether the sequence $9,-6,4,\dfrac{-8}{3},......$ is geometric and if it is, what is the common ratio?

Answer 1

Hint: We have been asked to verify whether the given sequence $9,-6,4,\dfrac{-8}{3},.....$ is geometric or not. From the basic concepts we know that the terms of the geometric sequence have a common ratio. So here we need to verify the ratio between different consecutive terms.

Complete step-by-step solution:
Now considering from the question we have been asked to verify whether the given sequence $9,-6,4,\dfrac{-8}{3},.....$ is geometric or not.
From the basic concepts of sequences we know that the consecutive terms of the geometric sequence have a common ratio.
So here we need to verify the ratio between different consecutive terms.
We will have a ratio of $9,-6$ , $-6,4$ and $4,\dfrac{-8}{3}$ . The ratio between $9,-6$ is $\Rightarrow \dfrac{-6}{9}=\dfrac{-2}{3}$ . Similarly the ratio between $-6,4$ is $\Rightarrow \dfrac{4}{-6}=\dfrac{2}{-3}$ . Now we will verify the ratio of $4,\dfrac{-8}{3}$ which is $\Rightarrow \dfrac{\left( \frac{-8}{3} \right)}{4}=\dfrac{-2}{3}$ .
If we observe all the ratios are equal. Therefore these consecutive terms have a common ratio.
Hence we can conclude that the given terms are in geometric sequence.

Note: This type of questions are very simple, involve less calculations, very few mistakes are possible and can be solved in a less span of time. We can also find the ${{n}^{th}}$ term of the sequence by using the formula given as $a{{r}^{n-1}}$ where $a$ is the first term and $r$ is the common ratio. For the given sequence the ${{n}^{th}}$ term will be given as $\Rightarrow \left( 9 \right){{\left( \dfrac{-2}{3} \right)}^{n-1}}$ .