If the 9th and 19th terms of a G.P. are 12 and 18, what is the G.M. of the first 27 terms of this G.P.?

Options

A. A)9
B. B)14
C. C)15
D. D)none of these

Answer

To solve the problem, we need to find the geometric mean (G.M.) of the first 27 terms of a geometric progression (G.P.) given that the 9th term is 12 and the 19th term is 18.

### Step-by-Step Solution:

1. **Understand the terms of a G.P.**:
In a G.P., the n-th term can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Write the equations for the 9th and 19th terms**:
From the problem, we have:
- 9th term: \( T_9 = a \cdot r^{8} = 12 \) (Equation 1)
- 19th term: \( T_{19} = a \cdot r^{18} = 18 \) (Equation 2)

3. **Divide Equation 2 by Equation 1**:
This will help eliminate \( a \):
\[ \frac{T_{19}}{T_9} = \frac{a \cdot r^{18}}{a \cdot r^{8}} = \frac{18}{12} \]
Simplifying this gives:
\[ r^{10} = \frac{18}{12} = \frac{3}{2} \]

4. **Solve for \( r \)**:
To find \( r \), take the 10th root of both sides:
\[ r = \left(\frac{3}{2}\right)^{\frac{1}{10}} \]

5. **Substitute \( r \) back to find \( a \)**:
Use Equation 1 to find \( a \):
\[ a \cdot r^{8} = 12 \]
Substitute \( r \):
\[ a \cdot \left(\left(\frac{3}{2}\right)^{\frac{1}{10}}\right)^{8} = 12 \]
Simplifying gives:
\[ a \cdot \left(\frac{3}{2}\right)^{\frac{8}{10}} = 12 \]
\[ a \cdot \left(\frac{3}{2}\right)^{0.8} = 12 \]
Thus,
\[ a = \frac{12}{\left(\frac{3}{2}\right)^{0.8}} \]

6. **Find the 27th term**:
The 27th term is given by:
\[ T_{27} = a \cdot r^{26} \]
Substitute \( a \) and \( r \):
\[ T_{27} = \frac{12}{\left(\frac{3}{2}\right)^{0.8}} \cdot \left(\frac{3}{2}\right)^{\frac{26}{10}} \]
Combine the powers of \( r \):
\[ T_{27} = \frac{12 \cdot \left(\frac{3}{2}\right)^{\frac{26}{10} - 0.8}}{1} \]
Simplifying gives:
\[ T_{27} = \frac{12 \cdot \left(\frac{3}{2}\right)^{\frac{26 - 8}{10}}}{1} = \frac{12 \cdot \left(\frac{3}{2}\right)^{\frac{18}{10}}} \]

7. **Calculate the G.M. of the first 27 terms**:
The geometric mean of the first \( n \) terms of a G.P. is given by:
\[ G.M. = \sqrt{T_1 \cdot T_{27}} \]
Since \( T_1 = a \):
\[ G.M. = \sqrt{a \cdot T_{27}} = \sqrt{a \cdot \left(\frac{12 \cdot \left(\frac{3}{2}\right)^{\frac{18}{10}}}{1}\right)} \]

After calculating, we find:
\[ G.M. = 6\sqrt{6} \]

### Final Answer:
The geometric mean of the first 27 terms of this G.P. is \( 6\sqrt{6} \).