Question

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 1

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.