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

### Step 1: Understand the formula for the nth term of a GP
The nth term \( T_n \) of a GP can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

### Step 2: Write the expression for the sixth term
Given that the sixth term \( T_6 \) is 2, we can write:
\[ T_6 = a \cdot r^{6-1} = a \cdot r^5 = 2 \]

### Step 3: Find the product of the first eleven terms
The product \( P \) of the first \( n \) terms of a GP can be expressed as:
\[ P = T_1 \cdot T_2 \cdot T_3 \cdots T_n = a \cdot (a \cdot r) \cdot (a \cdot r^2) \cdots (a \cdot r^{n-1}) \]
This can be simplified to:
\[ P = a^n \cdot r^{0 + 1 + 2 + \cdots + (n-1)} \]
The sum of the first \( n-1 \) integers is given by:
\[ \frac{(n-1) \cdot n}{2} \]
Thus, for \( n = 11 \):
\[ P = a^{11} \cdot r^{\frac{10 \cdot 11}{2}} = a^{11} \cdot r^{55} \]

### Step 4: Substitute \( a \) in terms of \( r \)
From the equation \( a \cdot r^5 = 2 \), we can express \( a \) as:
\[ a = \frac{2}{r^5} \]

### Step 5: Substitute \( a \) into the product equation
Now substituting \( a \) into the product expression:
\[ P = \left(\frac{2}{r^5}\right)^{11} \cdot r^{55} \]
This simplifies to:
\[ P = \frac{2^{11}}{r^{55}} \cdot r^{55} = 2^{11} \]

### Step 6: Calculate \( 2^{11} \)
Now we calculate \( 2^{11} \):
\[ 2^{11} = 2048 \]

### Final Answer
Thus, the product of the first eleven terms of the GP is:
\[ \boxed{2048} \]

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