Question:medium

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

Show Hint

The geometric mean of a GP's terms equals the middle term, found by averaging the exponents.
Updated On: Jul 21, 2026
  • \( 3^{\frac{n(n+1)}{2}} \)
  • \( 3^{\frac{n}{2}} \)
  • \( 3^{n} \)
  • \( 3^{2n} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Work with the exponents instead of the terms.
Each term is $3^k$ for $k = 0, 1, 2, \ldots, n$, so the exponents $0, 1, \ldots, n$ form an arithmetic progression.

Step 2: Take logs and average.
The log of the geometric mean equals the average of the logs of the terms, which equals $\log 3$ times the average of the exponents.
For an AP running from 0 to $n$, the average of all terms equals the average of the first and last terms, $\dfrac{0+n}{2} = \dfrac{n}{2}$.

Step 3: Convert back from logs.
So $\log(\text{GM}) = \dfrac{n}{2}\log 3$, which gives GM $= 3^{n/2}$.

Final Answer:
The geometric mean is $3^{n/2}$, matching option (b). \[ \boxed{3^{n/2}} \]
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