Question:medium

What is the geometric mean of the sequence 1, 3, 9, 27, 81,... 3n?

Updated On: Jul 20, 2026
  • \(3^{\frac{n(n+1)}{2}}\)
  • \(3^{\frac{n}{2}}\)
  • \(3^{n}\)
  • \(3^{2n}\)
  • \(3^{n(n+1)}\)
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The Correct Option is B

Solution and Explanation

The correct answer is option (B):
\(3^{\frac{n}{2}}\)

The geometric mean of a sequence is the nth root of the product of the terms, where n is the number of terms. In this case, we have the sequence 1, 3, 9, 27, 81,... 3^n. We need to first figure out how many terms are in the sequence. The sequence can be written as 3^0, 3^1, 3^2, 3^3, ... 3^n. The exponents range from 0 to n. To find the number of terms, we count from 0 to n, inclusive. So, there are n+1 terms in the sequence.

Now, to find the geometric mean, we first multiply all the terms together. This is:
1 * 3 * 9 * 27 * ... * 3^n = 3^0 * 3^1 * 3^2 * 3^3 * ... * 3^n.
When multiplying exponents with the same base, we add the powers. So the product becomes 3^(0 + 1 + 2 + 3 + ... + n).

The sum of integers from 0 to n is given by the formula n(n+1)/2. Therefore, the product of the terms is 3^(n(n+1)/2).

Finally, we take the (n+1)th root of the product to find the geometric mean. This means we calculate [3^(n(n+1)/2)]^(1/(n+1)). Using the properties of exponents, we multiply the exponents:

(n(n+1)/2) * (1/(n+1)) = n/2.

Therefore, the geometric mean is 3^(n/2).
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