Question

In a set of 2n observations the geometric mean of first 'n' observations is 81 and the geometric mean of remaining n-observations is 16 then the geometric mean of all 2n observations is

• A. 9
• B. 6
• C. 54
• D. 36

Hint:

If two groups have the same number of observations, their combined Geometric Mean is simply the Geometric Mean of their individual Geometric Means: $G = \sqrt{G_{1} \cdot G_{2}}$. This shortcut saves you from dealing with $n$ and $2n$ during the exam.

Answer 1

The Correct Option is D

This problem requires calculating the combined Geometric Mean (G.M.) of two groups of equal size.

Step 1: \color{redDefine the G.M. for each group
Let the first group of $n$ observations be $x_{1}, \dots, x_{n}$.
$G_{1} = (\prod_{i=1}^{n} x_{i})^{1/n} = 81$.
Let the second group of $n$ observations be $x_{n+1}, \dots, x_{2n}$.
$G_{2} = (\prod_{i=n+1}^{2n} x_{i})^{1/n} = 16$.

Step 2: \color{redCalculate the product of terms in each group
From the definitions in Step 1, we can find the total product of the observations:
Product of Group 1: $P_{1} = \prod_{i=1}^{n} x_{i} = (G_{1})^{n} = 81^{n}$.
Product of Group 2: $P_{2} = \prod_{i=n+1}^{2n} x_{i} = (G_{2})^{n} = 16^{n}$.

Step 3: \color{redFormulate the combined Geometric Mean
The G.M. of the entire set of $2n$ observations ($G_{total}$) is the $2n$-th root of the product of all terms:
$G_{total} = (P_{1} \cdot P_{2})^{1/(2n)}$.
Substituting the values from
Step 2:
$G_{total} = (81^{n} \cdot 16^{n})^{1/(2n)}$.

Step 4: \color{redSimplify the Power Expression
$G_{total} = [(81 \cdot 16)^{n}]^{1/(2n)}$.
$G_{total} = (81 \cdot 16)^{n/2n} = (81 \cdot 16)^{1/2}$.
This is equivalent to the square root of the product:
$G_{total} = \sqrt{81} \cdot \sqrt{16}$.
$G_{total} = 9 \cdot 4 = 36$.
The combined geometric mean is 36, which matches Option (4).