The 5th term of the series `(10)/(9),(1)/(3)sqrt((20)/(3)),(2)/(3),…` is

Options

A. `(1)/(3)`
B. 1
C. `(2)/(5)`
D. `sqrt((2)/(3))`

Answer

To find the fifth term of the series \( \frac{10}{9}, \frac{1}{3}\sqrt{\frac{20}{3}}, \frac{2}{3}, \ldots \), we will first determine if the series is a geometric progression (GP) and then use the properties of GPs to find the fifth term.

### Step 1: Identify the Terms
The first few terms of the series are:
- \( a_1 = \frac{10}{9} \)
- \( a_2 = \frac{1}{3}\sqrt{\frac{20}{3}} \)
- \( a_3 = \frac{2}{3} \)

### Step 2: Find the Common Ratio
To check if the series is a GP, we need to find the common ratio \( r \) by calculating the ratio of consecutive terms.

**Finding \( r_1 = \frac{a_2}{a_1} \)**:
\[ r_1 = \frac{\frac{1}{3}\sqrt{\frac{20}{3}}}{\frac{10}{9}} = \frac{1}{3}\sqrt{\frac{20}{3}} \cdot \frac{9}{10} \]
Simplifying this:
\[ = \frac{9}{30}\sqrt{\frac{20}{3}} = \frac{3}{10}\sqrt{\frac{20}{3}} = \frac{3\sqrt{20}}{10\sqrt{3}} = \frac{3\sqrt{20}}{10\sqrt{3}} = \frac{3\sqrt{20}}{10\sqrt{3}} = \frac{3\cdot 2\sqrt{5}}{10\sqrt{3}} = \frac{6\sqrt{5}}{10\sqrt{3}} = \frac{3\sqrt{5}}{5\sqrt{3}} \]

**Finding \( r_2 = \frac{a_3}{a_2} \)**:
\[ r_2 = \frac{\frac{2}{3}}{\frac{1}{3}\sqrt{\frac{20}{3}}} = \frac{2}{3} \cdot \frac{3}{\sqrt{\frac{20}{3}}} = \frac{2}{\sqrt{\frac{20}{3}}} = \frac{2\sqrt{3}}{\sqrt{20}} = \frac{2\sqrt{3}}{2\sqrt{5}} = \frac{\sqrt{3}}{\sqrt{5}} \]

### Step 3: Check if the Ratios are Equal
We need to check if \( r_1 = r_2 \):
\[ r_1 = \frac{3\sqrt{5}}{5\sqrt{3}}, \quad r_2 = \frac{\sqrt{3}}{\sqrt{5}} \]
Cross-multiplying to check equality:
\[ 3\sqrt{5} \cdot \sqrt{5} = 5\sqrt{3} \cdot \sqrt{3} \implies 15 = 15 \quad \text{(True)} \]
Thus, the series is a GP.

### Step 4: Find the Common Ratio
We can simplify the common ratio:
\[ r = \frac{\sqrt{6}}{\sqrt{10}} = \sqrt{\frac{6}{10}} = \sqrt{\frac{3}{5}} \]

### Step 5: Find the Fifth Term
The \( n \)-th term of a GP is given by:
\[ a_n = a_1 \cdot r^{n-1} \]
For the fifth term \( a_5 \):
\[ a_5 = a_1 \cdot r^{4} = \frac{10}{9} \cdot \left(\sqrt{\frac{3}{5}}\right)^{4} \]
Calculating \( r^4 \):
\[ r^4 = \left(\frac{3}{5}\right)^{2} = \frac{9}{25} \]
Thus,
\[ a_5 = \frac{10}{9} \cdot \frac{9}{25} = \frac{10}{25} = \frac{2}{5} \]

### Final Answer
The fifth term of the series is \( \frac{2}{5} \).

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