Question

If ${{A}_{1}},{{A}_{2}}$ be two arithmetic means and ${{G}_{1}},{{G}_{2}}$ be two geometric means between two positive numbers $a$ and $b$, then $\dfrac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}$
$A)\dfrac{ab}{a+b}$
$B)\dfrac{ab}{a+b}$
$C)\dfrac{a+b}{ab}$
$D)\dfrac{a+b}{ab}$

Answer 1

Hint: To solve this question we need to know about the Arithmetic means and geometric means. To solve the problem we will take an assumption to find the geometric means ${{G}_{1}},{{G}_{2}}$and will then calculate for ${{A}_{1}},{{A}_{2}}$ .

Complete step by step answer:
The question ask us to find the value for $\dfrac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}$, if ${{A}_{1}},{{A}_{2}}$ be two arithmetic means and ${{G}_{1}},{{G}_{2}}$ be two geometric means between two positive numbers $a$ and $b$. To solve this problem we first find the geometric means and the arithmetic mean. For solving the geometric mean we will make an assumption to make the solving easier, where we will consider $a={{p}^{3}}$ and $b={{q}^{3}}$. The product of the geometric mean when a series will look like $a,{{G}_{1}},{{G}_{2}},b$ which will as per the assumption be${{p}^{3}},{{G}_{1}},{{G}_{2}},{{q}^{3}}$. So ${{G}_{1}}={{p}^{2}}q$ while ${{G}_{1}}=p{{q}^{2}}$. So the product of the two means will be:
$\Rightarrow {{G}_{1}}{{G}_{2}}={{p}^{2}}q\times p{{q}^{2}}$
On multiplying the expression we get:
$\Rightarrow {{G}_{1}}{{G}_{2}}={{p}^{3}}{{q}^{3}}$
On substituting $p,q$ from $a$ and $b$, we get:
$\Rightarrow {{G}_{1}}{{G}_{2}}=ab$

The second step is to find the value for the arithmetic mean expression. The series is $a,{{A}_{1}},{{A}_{2}},b$. Now considering the $b$ to be the last term and $a$ to be the first term. The value of $b$ in terms of $a$ will be:
$\Rightarrow b=a+3d$
We will rearrange the expression to calculate for $d$. On doing this we get:
$\Rightarrow b=a+3d$
$\Rightarrow d=\dfrac{b-a}{3}$
Now on applying the same formula to find the value for ${{A}_{1}},{{A}_{2}}$ in terms of $a$ and $b$we get:
$\Rightarrow {{A}_{1}}=a+d$
On substituting the value of $d$ we get:
$\Rightarrow {{A}_{1}}=a+\dfrac{b-a}{3}$
$\Rightarrow {{A}_{1}}=\dfrac{3a+b-a}{3}$
$\Rightarrow {{A}_{1}}=\dfrac{2a+b}{3}$
In the similar manner we will find the value of ${{A}_{2}}$. The formula will be:
$\Rightarrow {{A}_{2}}=a+2d$
On substituting the value of $d$ we get:
$\Rightarrow {{A}_{2}}=a+2\left( \dfrac{b-a}{3} \right)$
$\Rightarrow {{A}_{2}}=\dfrac{3a+2b-2a}{3}$
$\Rightarrow {{A}_{2}}=\dfrac{a+2b}{3}$
The value of the sum of arithmetic mean will be:
 \[\Rightarrow {{A}_{1}}+{{A}_{2}}=\dfrac{2a+b}{3}+\dfrac{a+2b}{3}\]
\[\Rightarrow {{A}_{1}}+{{A}_{2}}=\dfrac{3a+3b}{3}\]
\[\Rightarrow {{A}_{1}}+{{A}_{2}}=a+b\]
Now the value of the expression $\dfrac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}$will be:
$\Rightarrow \dfrac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}=\dfrac{a+b}{ab}$
$\therefore $ The value of $\dfrac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}$is $C)\dfrac{a+b}{ab}$.

So, the correct answer is “Option C”.

Note: We need to remember when we talk of the mean for a particular series then the mean is also considered as the part of the series. For example if ${{A}_{1}},{{A}_{2}}{{A}_{3}},{{A}_{4}}$ are the arithmetic means between $a$ and $b$, then $a,{{A}_{1}},{{A}_{2}}{{A}_{3}},{{A}_{4}},b$ is in the arithmetic progression. Similar rule lies for the geometric mean also.