# 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. \[\frac{{ab}}{{a + b}}\]

Answer

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**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.

Last updated date: 06th Jun 2023

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