Chapter 1, Problem 3MCQ

Solution

Find the series of $\frac{2x}{3},\frac{x}{3},\frac{x}{6},\frac{x}{12}\mathrm{............},$

The value is $\frac{x}{24}$ .

Hence, the series is $\frac{2x}{3},\frac{x}{3},\frac{x}{6},\frac{2x}{3},\underset{¯}{\frac{x}{24}}$

Given:

The Geometric series: $\frac{2x}{3},\frac{x}{3},\frac{x}{6},\frac{x}{12}\mathrm{............},$

Calculation:

Concept Used:

For any geometric progression, the $n\text{th}$ term is defined is as follows:

  $a_n=a{r}^{n-1}$

Where, $a$ is the first time of the arithmetic progression.

  $r=\frac{a_i}{a_{i-1}}$ is the common ratio of the arithmetic progression.

Calculation:

In the given arithmetic progression, $\frac{2x}{3},\frac{x}{3},\frac{x}{6},\frac{x}{12}\mathrm{............},$

  $a=-405$ is the first time of the arithmetic progression.

  $r=\frac{a_i}{a_{i-1}}$ is the common ratio of two consecutive term of the geometric progression which is calculated as follows:

  $\begin{array}{l}r=\frac{a_i}{a_{i-1}}\\ r=\frac{a_2}{a_1}\left[\text{take,}\text{i=2}\right]\\ r=\frac{\frac{x}{3}}{2\frac{x}{3}}\\ r=\frac{1}{2}\end{array}$

Here the required next term is $5\text{th}$ term of the sequence.

Hence the required term is calculated as follow:

  $\begin{array}{l}{a}_n=a{r}^{n-1}\\ a_5=2\frac{x}{3}\times {\left(\frac{1}{2}\right)}^{5-1}\\ a_5=\frac{x}{3\times 2\times 2\times 2}\\ a_5=\frac{x}{24}\end{array}$

Hence $5\text{th}$ term is $a_5=\frac{x}{24}$

Conclusion:

The required next term is $\frac{x}{24}$