Question

If $6,{6^2},{6^3},{6^4}{..........6^n}$ be n observation, then the quantity ${\left( {\sqrt 6 } \right)^{n + 1}}$ is called

$$A.{\text{ G}}{\text{.M}}{\text{.}} \\\ B.{\text{ H}}{\text{.M}}{\text{.}} \\\ C.{\text{ A}}{\text{.M}}{\text{.}} \\\ D.{\text{ None of these}} \\\$$

Answer 1

Hint: In order to solve such questions with given value and missing terms the easiest way is to check out all the options available for selection of correct. Also try to identify the series.

Complete step-by-step answer:
Given series: $6,{6^2},{6^3},{6^4}{..........6^n}$
Since, in the above series each term after the other term changes by the product of 6 from the last one. This is the characteristic of geometric progression series. So most probably the answer will be found out easily if we start with G.M.
Checking for option (A): G.M.
As we know that for a general geometric series with terms $x,{x^2},{x^3},{x^4}..........{x^n}$ . The geometric mean of the series is given by:
${\bar X_{geom}} = \sqrt[n]{{x.{x^2}.{x^3}.{x^4}..........{x^n}}} = {\left( {x.{x^2}.{x^3}.{x^4}..........{x^n}} \right)^{\dfrac{1}{n}}}$
So, the geometric mean of the given series is given by:
${{\bar X}_{geom}} = \sqrt[n]{{{{6.6}^2}{{.6}^3}{{.6}^4}{{..........6}^n}}} \\\ = {\left( {{{6.6}^2}{{.6}^3}{{.6}^4}{{..........6}^n}} \right)^{\dfrac{1}{n}}} \\\ = {\left( {{6^{1 + 2 + 3 + 4 + ........ + n}}} \right)^{\dfrac{1}{n}}} \\\$
The power of 6 in the internal bracket is the sum of n consecutive numbers which is an A.P.
$\Rightarrow 1 + 2 + 3 + 4 + ........ + n = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {{\text{sum of n numbers of A}}{\text{.P}}{\text{.}}} \right]$
So, the geometric mean becomes

$${{\bar X}_{geom}} = {\left( {{6^{1 + 2 + 3 + 4 + ........ + n}}} \right)^{\dfrac{1}{n}}} \\\ {{\bar X}_{geom}} = {\left( {{6^{\dfrac{{n\left( {n + 1} \right)}}{2}}}} \right)^{\dfrac{1}{n}}} = {\left( 6 \right)^{\dfrac{{n\left( {n + 1} \right)}}{2} \times \dfrac{1}{n}}}{\text{ }}\left[ {\because {{\left( {{x^a}} \right)}^b} = {{\left( x \right)}^{a \times b}}} \right] \\\ = {\left( 6 \right)^{\dfrac{{\left( {n + 1} \right)}}{2}}} \\\ = {\left( {{6^{\dfrac{1}{2}}}} \right)^{\left( {n + 1} \right)}} = {\left( {\sqrt 6 } \right)^{\left( {n + 1} \right)}} \\\$$

Hence, the geometric mean of the numbers is the same as the quantity given.
So, option A is the correct option.

Note: In order to solve such problems try to use a direct formula for the mean of different numbers. The geometric mean (or GM) is a type of mean that indicates the central tendency of a set of numbers by using the product of their values. It is defined as the nth root of the product of n numbers. Remember the formulas for the mean and sum of n numbers in Geometric progression and also in Arithmetic progression.