Question

The G.M of the numbers $ 3,{3^2},{3^3}.....,{3^n} $ is
 $ \eqalign{
  & 1){3^{\dfrac{2}{n}}} \cr
  & 2){3^{\dfrac{{\left( {n + 1} \right)}}{2}}} \cr
  & 3){3^{\dfrac{n}{2}}} \cr
  & 4){3^{\dfrac{{\left( {n - 1} \right)}}{2}}} \cr} $

Answer 1

Hint: The given expression is in the form of a Geometric Progression. In a geometric sequence is a sequence where each term after the first one is found by multiplying the previous one. The question asks for G.M which means the Geometric Mean of the given sequence.
The formula used to solve this problem is:
 $ G.M = {\left( {{x_1} \times {x_2} \times {x_3} \times ......{x_n}} \right)^{\dfrac{1}{n}}} $
Where, $ {x_1},{x_2},....{x_n} $ are the observations.

Complete step-by-step answer:
The given sequence in the above problem is, $ 3,{3^2},{3^3}.....,{3^n} $
Using the above formula, the Geometric Mean will be,
 $ G.M = {\left( {3 \times {3^2} \times {3^3}.....,{3^n}} \right)^{\dfrac{1}{n}}} $
Since the base is equal, the powers can be added.
 $ G.M = {\left( {{3^{1 + 2 + 3 + ..... + n}}} \right)^{\dfrac{1}{n}}} $
Now the expression, $ {3^{1 + 2 + 3 + ..... + n}} $ can be written as, $ {3^{\dfrac{{n\left( {n + 1} \right)}}{2}}} $
Substituting the value,
 $ G.M = {\left( {{3^{\dfrac{{n\left( {n + 1} \right)}}{2}}}} \right)^{\dfrac{1}{n}}} $
Now, we can multiply the power outside the bracket with the power inside the bracket.
 $ G.M = {3^{\dfrac{{\left( {n + 1} \right)}}{2}}} $
Therefore, the final answer is $ {3^{\dfrac{{\left( {n + 1} \right)}}{2}}} $ .
Hence, option (2) is the correct answer.
So, the correct answer is “Option 2”.

Note: Geometric mean is a special type of average where we multiply the numbers together and take the roots of the number of powers present in the sequence. Geometric Mean is a type of mean that indicates the central tendency of a set of numbers by using the product of their values.