Question

If $a,b,c$ are in A.P., $x$ is the G.M. between $a$ and $b$, $y$ is the G.M. between $b$ and $c$, then ${b^2}$ is equal to
A.$\dfrac{1}{2}\left( {{x^2} + {y^2}} \right)$
B.$xy$
C.$\dfrac{{2{x^2}{y^2}}}{{{x^2} + {y^2}}}$
D.None of these

Answer 1

Hint: Here, we will find the square of the variable. First, we will use the arithmetic progression formula to find the equation. Then we will use the geometric mean formula from the given to find the square of the variable. Arithmetic Progression is a sequence of numbers such that the common difference between any two consecutive numbers is a constant.

Formula Used:
We will use the following formula:
1.In an Arithmetic Progression, Common difference is given by the formula $d = {a_{n + 1}} - {a_n}$
2.Geometric mean is given by the formula $\overline {{X_{geo}}} = \sqrt[n]{{{x_1}{x_2}{x_3}......{x_n}}}$

Complete step-by-step answer:
We are given that $a,b,c$ are in A.P.
We know that in Arithmetic Progression, the common difference is the same for all the terms. So, we get
${d_1} = {d_2}$
Then using the formula $d = {a_{n + 1}} - {a_n}$, we get
$ \Rightarrow b - a = c - b$
By rewriting the equation, we get
$ \Rightarrow b + b = a + c$
By adding the like terms, we get
$ \Rightarrow 2b = a + c$ …………………………………………………………..$\left( 1 \right)$
We are given that $x$ is the G.M. between $a$ and $b$.
Substituting the values in the formula $\overline {{X_{geo}}} = \sqrt[n]{{{x_1}{x_2}{x_3}......{x_n}}}$, we get
$x = \sqrt {a \cdot b} $
$ \Rightarrow x = \sqrt {ab} $
By squaring on both the sides, we get
$ \Rightarrow {x^2} = ab$ ……………………………………………………………..$\left( 2 \right)$
We are given that $y$ is the G.M. between $b$ and $c$.
Substituting the values in the formula $\overline {{X_{geo}}} = \sqrt[n]{{{x_1}{x_2}{x_3}......{x_n}}}$, we get
$y = \sqrt {b \cdot c} $
$ \Rightarrow y = \sqrt {bc} $
By squaring on both the sides, we get
$ \Rightarrow {y^2} = bc$ ………………………………………………………………$\left( 3 \right)$
By adding equations $\left( 2 \right)$ and $\left( 3 \right)$, we get
${x^2} + {y^2} = ab + bc$
By taking out common factors, we get
$ \Rightarrow {x^2} + {y^2} = b\left( {a + c} \right)$
By substituting equation $\left( 1 \right)$, we get
$ \Rightarrow {x^2} + {y^2} = b\left( {2b} \right)$
$ \Rightarrow {x^2} + {y^2} = 2{b^2}$
By rewriting the equation, we get
$ \Rightarrow {b^2} = \dfrac{1}{2}\left( {{x^2} + {y^2}} \right)$
Thus ${b^2}$ is the arithmetic mean of the sum of the geometric mean of the terms.
Therefore ${b^2}$ is equal to $\dfrac{1}{2}\left( {{x^2} + {y^2}} \right)$.
Thus, option (A) is the correct answer.

Note: We should note that the geometric mean is defined as the ${n^{th}}$ root of the product of $n$ variables. We are provided with the geometric mean of only two variables, so we should take the square root of the product of the variables. We might make a mistake by using the arithmetic mean formula instead of the geometric mean. The arithmetic mean is defined as the average of a particular set of numbers and is calculated by adding all the numbers given and then dividing it by the total number of observations.