Question

If the $6^{\text{th}}$ term of a G.P. is $2$, then the product of the first $11$ terms of the G.P. is equal to:

• A. \( 512 \)
• B. \( 1024 \)
• C. \( 2048 \)
• D. \( 256 \)
• E. \( 32 \)

Hint:

For any G.P. with \( n \) terms, the product of all terms is \( (\text{middle term})^n \), provided \( n \) is odd. Since 11 is odd, the middle term is the \( \frac{11+1}{2} = 6^{th} \) term. Hence, \( \text{Product} = (a_6)^{11} \).

Answer 1

The Correct Option is C

Concept: The general term of a Geometric Progression (G.P.) is defined as \( a_n = ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. A useful property of G.P. is that the product of terms equidistant from the beginning and the end is constant and equal to the square of the middle term (if one exists).

Step 1: Define the given information mathematically.
We are told the \( 6^{th} \) term is 2: \[ a_6 = ar^{6-1} = ar^5 = 2 \quad \cdots (1) \]

Step 2: Write the expression for the product of the first 11 terms.
Let the product be \( P \): \[ P = a_1 \times a_2 \times a_3 \times \ldots \times a_{11} \] \[ P = (a) \times (ar) \times (ar^2) \times \ldots \times (ar^{10}) \]

Step 3: Simplify the product using laws of exponents.
Grouping the terms: \[ P = a^{11} \cdot r^{(0 + 1 + 2 + \ldots + 10)} \] The sum of the first 10 integers is calculated using \( \frac{n(n+1)}{2} \): \[ \text{Sum} = \frac{10(11)}{2} = 55 \] Thus, the product becomes: \[ P = a^{11} r^{55} \]

Step 4: Relate the product to the given \( 6^{th} \) term.
Observe that \( a^{11} r^{55} \) can be rewritten as a power of the \( 6^{th} \) term: \[ P = (ar^5)^{11} \] Substitute the value from equation (1): \[ P = (2)^{11} \] Calculating the power: \[ 2^{10} = 1024 \Rightarrow 2^{11} = 2048 \]