Question

If the geometric mean of the numbers 40, 50 and x is 10, then the value of x is
[a] $\dfrac{1}{2}$
[b] 4
[c] 6
[d] 2

Answer 1

Hint: Geometric mean of n observation ${{a}_{1}},{{a}_{2}},\cdots ,{{a}_{n}}$ is given by ${{\left( \prod\limits_{r=1}^{n}{{{a}_{r}}} \right)}^{\dfrac{1}{n}}}$. Use the above result to get an equation in x. Solve the equation for x. Hence find the value of x.



Complete step-by-step answer:
Here we have three terms a = 40, b = 50 and x.
We know that Geometric mean of n observation ${{a}_{1}},{{a}_{2}},\cdots ,{{a}_{n}}$ is given by ${{\left( \prod\limits_{r=1}^{n}{{{a}_{r}}} \right)}^{\dfrac{1}{n}}}$
Hence the geometric mean of three terms a, b and c is given by ${{\left( abc \right)}^{\dfrac{1}{3}}}$
But given that the geometric mean is 10.
Hence, we have $10={{\left( 40\times 50x \right)}^{\dfrac{1}{3}}}$
Cubing both sides, we get
${{10}^{3}}=2000x$
Dividing both sides by 2000, we get
$x=\dfrac{1000}{2000}$
i.e. $x=\dfrac{1}{2}$
Hence the value of x is 0.5
Hence option [a] is correct.

Note: [1] In geometric means, we use the product of numbers as a measure of central tendency of a given sequence of numbers.
[2] In arithmetic mean, we use the sum of the numbers as a measure of the central tendency of a given sequence of numbers. The arithmetic mean of n numbers is given by $\dfrac{\sum\limits_{r=1}^{n}{{{a}_{r}}}}{n}$.
[3] If all the numbers in distribution are positive, then $\text{A}\text{.M}\ge \text{G}\text{.M}$ and the equality holds when the numbers in the distribution are all equal. This is called AM-GM inequality and is one of the most widely used inequalities in mathematics.