Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If ${A_1},{A_2}$ are the two A.M.'s between two numbers $a$ and $b$ ${G_1},{G_2}$ be two G.M.'s between same two numbers, then $\frac{{{A_1} + {A_2}}}{{{G_1}{\rm{. }}{G_2}}}$A. $\frac{{a + b}}{{ab}}$В. $\frac{{a + b}}{{2ab}}$C. $\frac{{2ab}}{{a + b}}$D. $\frac{{ab}}{{a + b}}$

Last updated date: 13th Jul 2024
Total views: 60.9k
Views today: 0.60k
Verified
60.9k+ views
Hint:
To answer this question, we must first understand the Arithmetic and Geometric Means. To solve the problem, we will make an assumption to determine the geometric means $G1,G2$, and then calculate $A1,A2$. To begin solving this problem, we must first determine the geometric and arithmetic means. In order to solve the geometric mean, we will make an assumption in which $a = {p^3}$and $b = {q^3}$are considered.
Formula used:
The A.M of a,b term of A.P is $\frac{{(a + b)}}{2}$ )
The G.M of a,b term of G.P is $\sqrt {ab}$
The H.M of a, b term of H.P is $\frac{{2ab}}{{(a + b)}}$
Complete step-by-step solution:
The question asks us to calculate $\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}}$ given two arithmetic means and two geometric means between two positive numbers ‘a’ and ‘b’.
Now, we have to solve product of the two means, we obtain
$\Rightarrow {G_1}{G_2} = {p^2}q \times p{q^2}$
Now, by multiplying the above expression we get:
$\Rightarrow {G_1}{G_2} = {p^3}{q^3}$
Here, we have to substitute $p,q$ from $a$ and $b$, we get:
$\Rightarrow {G_1}{G_2} = ab$
The next step is to calculate the value of the arithmetic mean expression.
The sequence is as follows:
$a,{A_1},{A_2},b$.
Considering that b is the last term and “a” is the first term. In terms of a, the value of b will be
$b = a + 3d$
To calculate for$d$, we have to rearrange the expression:
On doing this we get:
$\Rightarrow b = a + 3d$
$\Rightarrow d = \frac{{b - a}}{3}$
To find the value for ${A_1},{A_2}$ in terms of $a$ and $b$, we have to apply the same formula we get:
$\Rightarrow {A_1} = a + d$
We have to substitute the value of $d$, we get
$\Rightarrow {A_1} = a + \frac{{b - a}}{3}$
$\Rightarrow {A_1} = \frac{{3a + b - a}}{3}$
$\Rightarrow {A_1} = \frac{{2a + b}}{3}$
In the similar manner we will find the value of ${A_2}$.
The formula will be:
$\Rightarrow {A_2} = a + 2d$
Here, on substituting the value of$d$we get:
$\Rightarrow {A_2} = a + 2\left( {\frac{{b - a}}{3}} \right)$
Make the denominator common for all the terms:
$\Rightarrow {A_2} = \frac{{3a + 2b - 2a}}{3}$
Now, simplify:
$\Rightarrow {A_2} = \frac{{a + 2b}}{3}$
The sum of arithmetic means will have the following value:
$\Rightarrow {A_1} + {A_2} = \frac{{2a + b}}{3} + \frac{{a + 2b}}{3}$
Now, make the denominator common by solving the numerator
$\Rightarrow {A_1} + {A_2} = \frac{{3a + 3b}}{3}$
Simplify the above expression by taking ${\rm{3}}$as common in the numerator and cancel in with denominator:
$\Rightarrow {A_1} + {A_2} = a + b$
The value of the expression $\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}}$ becomes,
$\Rightarrow \frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}} = \frac{{a + b}}{{ab}}$
Therefore, the value of$\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}}$is$\frac{{a + b}}{{ab}}$
Hence, the option A is correct.

Note:
We must remember that when we discuss the mean for a specific series, the mean is also considered to be a part of the series. For example, if ${A_1},{A_2},{A_3},{A_4},$ are the arithmetic means between $a$ and $b$, then the arithmetic progression is $a,{A_1},{A_2},{A_3},{A_4},b$. The geometric mean follows a similar rule.