Question

What is the geometric mean of the sequence \( 1, 3, 9, 27, 81, \ldots 3^{n} \)?

• A. \( 3^{\frac{n(n+1)}{2}} \)
• B. \( 3^{\frac{n}{2}} \)
• C. \( 3^{n} \)
• D. \( 3^{2n} \)

Hint:

The geometric mean of a GP's terms equals the middle term, found by averaging the exponents.

Answer 1

The Correct Option is B

Step 1: List the terms of the sequence.
The sequence runs \( 3^0, 3^1, 3^2, \ldots, 3^n \), which is \( n+1 \) terms in all.

Step 2: Multiply all the terms.
The product of the terms is \( 3^{0+1+2+\cdots+n} = 3^{\frac{n(n+1)}{2}} \), using the sum of the first \( n \) natural numbers.

Step 3: Take the \( (n+1) \)th root for the geometric mean.
GM \( = \left(3^{\frac{n(n+1)}{2}}\right)^{\frac{1}{n+1}} = 3^{\frac{n(n+1)}{2(n+1)}} = 3^{\frac{n}{2}} \).

Final Answer:
The geometric mean of the sequence is \( 3^{\frac{n}{2}} \). \[ \boxed{3^{\frac{n}{2}}} \]