If the sixth term of a GP be 2, then the product of
first eleven terms is

Options

A. 1024
B. 2047
C. 2048
D. 1023

Answer

To solve the problem, we need to find the product of the first eleven terms of a geometric progression (GP) given that the sixth term is equal to 2.

### Step-by-Step Solution:

1. **Understanding the Terms of GP**:
The terms of a GP can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)
- Sixth term: \( ar^5 \)
- Seventh term: \( ar^6 \)
- Eighth term: \( ar^7 \)
- Ninth term: \( ar^8 \)
- Tenth term: \( ar^9 \)
- Eleventh term: \( ar^{10} \)

2. **Finding the Sixth Term**:
We are given that the sixth term \( ar^5 = 2 \).

3. **Product of the First Eleven Terms**:
The product of the first eleven terms can be expressed as:
\[ P = a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 \cdot ar^5 \cdot ar^6 \cdot ar^7 \cdot ar^8 \cdot ar^9 \cdot ar^{10} \]
This simplifies to:
\[ P = a^{11} \cdot r^{0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10} \]
The sum of the exponents of \( r \) is:
\[ 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = \frac{10 \cdot 11}{2} = 55 \]
Thus, we have:
\[ P = a^{11} \cdot r^{55} \]

4. **Expressing \( r \) in Terms of \( a \)**:
From the sixth term \( ar^5 = 2 \), we can express \( r^5 \) as:
\[ r^5 = \frac{2}{a} \]
Therefore, we can find \( r \):
\[ r = \left(\frac{2}{a}\right)^{\frac{1}{5}} \]

5. **Substituting \( r \) Back into the Product**:
Now substituting \( r \) back into the product:
\[ P = a^{11} \cdot \left(\left(\frac{2}{a}\right)^{\frac{1}{5}}\right)^{55} \]
This simplifies to:
\[ P = a^{11} \cdot \left(\frac{2}{a}\right)^{11} = a^{11} \cdot \frac{2^{11}}{a^{11}} = 2^{11} \]

6. **Calculating \( 2^{11} \)**:
Finally, we calculate \( 2^{11} \):
\[ 2^{11} = 2048 \]

### Conclusion:
The product of the first eleven terms of the GP is \( 2048 \).