If `0.bar(27), x` and `0.bar(72)` are in H.P. , then x must be :

Options

A. Rational
B. Integer
C. Irrational
D. Natural number

Answer

To solve the problem, we need to find the value of \( x \) such that the numbers \( 0.\overline{27}, x, 0.\overline{72} \) are in Harmonic Progression (H.P.).

### Step-by-step Solution:

1. **Convert the repeating decimals to fractions:**
- Let \( y = 0.\overline{27} \).
- To convert \( y \) into a fraction, we can set up the equation:
\[ y = 0.272727\ldots \]
Multiplying both sides by 100:
\[ 100y = 27.272727\ldots \]
Now, subtract the first equation from the second:
\[ 100y - y = 27.272727\ldots - 0.272727\ldots \]
This simplifies to:
\[ 99y = 27 \implies y = \frac{27}{99} = \frac{3}{11} \]

2. **Convert the other repeating decimal:**
- Let \( z = 0.\overline{72} \).
- Similarly, set up the equation:
\[ z = 0.727272\ldots \]
Multiplying both sides by 100:
\[ 100z = 72.727272\ldots \]
Subtracting the first equation from the second:
\[ 100z - z = 72.727272\ldots - 0.727272\ldots \]
This simplifies to:
\[ 99z = 72 \implies z = \frac{72}{99} = \frac{8}{11} \]

3. **Using the property of Harmonic Progression:**
- For three numbers \( a, b, c \) to be in H.P., the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) must be in A.P. (Arithmetic Progression).
- Therefore, we have:
\[ \frac{1}{y}, x, \frac{1}{z} \]
- The condition for A.P. is:
\[ 2x = \frac{1}{y} + \frac{1}{z} \]
- Substituting the values of \( y \) and \( z \):
\[ 2x = \frac{1}{\frac{3}{11}} + \frac{1}{\frac{8}{11}} = \frac{11}{3} + \frac{11}{8} \]

4. **Finding a common denominator:**
- The common denominator of 3 and 8 is 24:
\[ \frac{11}{3} = \frac{88}{24}, \quad \frac{11}{8} = \frac{33}{24} \]
- Therefore:
\[ 2x = \frac{88}{24} + \frac{33}{24} = \frac{121}{24} \]

5. **Solving for \( x \):**
- Now, divide both sides by 2:
\[ x = \frac{121}{24 \cdot 2} = \frac{121}{48} \]

### Final Answer:
Thus, the value of \( x \) is \( \frac{121}{48} \).