Question

If ${{G}_{1}}$ and ${{G}_{2}}$ are geometric mean of two series of sizes ${{n}_{1}}$ and ${{n}_{2}}$ respectively and G is geometric mean of their combined series, then log G is equal to
A. $\log {{G}_{1}}+\log {{G}_{2}}$
B. ${{n}_{1}}\log {{G}_{1}}+{{n}_{2}}\log {{G}_{2}}$
C. $\dfrac{\log {{G}_{1}}+\log {{G}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
D. $\dfrac{{{n}_{1}}\log {{G}_{1}}+{{n}_{2}}\log {{G}_{2}}}{{{n}_{1}}+{{n}_{2}}}$

Answer 1

Hint: First of all, we have assume that ${{G}_{1}}$ is a series containing $({{x}_{1}},{{x}_{2}},....{{x}_{{{n}_{1}}}})$ and ${{G}_{2}}$ is a series containing $({{y}_{1}},{{y}_{2}},....{{y}_{{{n}_{2}}}})$. We have to use the equation to find the geometric mean. That is, ${{G}_{1}}={{({{x}_{1}},{{x}_{2}},....,{{x}_{{{n}_{1}}}})}^{\dfrac{1}{{{n}_{1}}}}}$ and similarly for ${{G}_{2}}$. It is also given that G is the geometric mean of their combined series. So, the equation will be $G={{({{x}_{1}},{{x}_{2}},...,{{x}_{{{n}_{1}}}})}^{\dfrac{1}{{{n}_{1}}}}}\times {{({{y}_{1}},{{y}_{2}},...,{{y}_{{{n}_{2}}}})}^{\dfrac{1}{{{n}_{2}}}}}$. After that we can find the value of log G.

Complete step-by-step answer:
First of all, we have assume that ${{G}_{1}}$ is a series containing $({{x}_{1}},{{x}_{2}},....{{x}_{{{n}_{1}}}})$ and ${{G}_{2}}$ is a series containing $({{y}_{1}},{{y}_{2}},....{{y}_{{{n}_{2}}}})$. Here, the size of ${{G}_{1}}$ is ${{n}_{1}}$ and ${{G}_{2}}$ is ${{n}_{2}}$.
Now, we can use the geometric mean, ${{G}_{1}}$.
${{G}_{1}}={{({{x}_{1}},{{x}_{2}},....,{{x}_{{{n}_{1}}}})}^{\dfrac{1}{{{n}_{1}}}}}$
$\Rightarrow $$({{x}_{1}},{{x}_{2}},....{{x}_{{{n}_{1}}}})=G_{1}^{{{n}_{1}}}$
The geometric mean, ${{G}_{2}}$ will be,
${{G}_{2}}={{({{y}_{1}},{{y}_{2}},....,{{y}_{{{n}_{2}}}})}^{\dfrac{1}{{{n}_{2}}}}}$
$\Rightarrow $$({{y}_{1}},{{y}_{2}},....{{y}_{{{n}_{2}}}})=G_{2}^{{{n}_{2}}}$
Now, it is said that geometric mean G is the combined mean of their series. So,
$G={{({{x}_{1}},{{x}_{2}},...,{{x}_{{{n}_{1}}}})}^{\dfrac{1}{{{n}_{1}}}}}\times {{({{y}_{1}},{{y}_{2}},...,{{y}_{{{n}_{2}}}})}^{\dfrac{1}{{{n}_{2}}}}}$
On simplifying, we get,
$G={{(({{x}_{1}},{{x}_{2}},...,{{x}_{{{n}_{1}}}})\times ({{y}_{1}},{{y}_{2}},...,{{y}_{{{n}_{2}}}}))}^{\dfrac{1}{{{n}_{1}}+{{n}_{2}}}}}$
Now, we can rewrite it as,
$G={{(G_{1}^{{{n}_{1}}}\times G_{2}^{{{n}_{2}}})}^{\dfrac{1}{{{n}_{1}}+{{n}_{2}}}}}$
Now, we can find the log G. So, the above equation becomes,
$\log G=\log {{(G_{1}^{{{n}_{1}}}\times G_{2}^{{{n}_{2}}})}^{\dfrac{1}{{{n}_{1}}+{{n}_{2}}}}}$
On solving, we get,
$\log G=\dfrac{\log (G_{1}^{{{n}_{1}}}\times G_{2}^{{{n}_{2}}})}{{{n}_{1}}+{{n}_{2}}}$
On further expanding, we get,
$\log G=\dfrac{\log (G_{1}^{{{n}_{1}}})+\log (G_{2}^{{{n}_{2}}})}{{{n}_{1}}+{{n}_{2}}}$
On further solving, we will get,
$\therefore \log G=\dfrac{{{n}_{1}}\log ({{G}_{1}})+{{n}_{2}}\log ({{G}_{2}})}{{{n}_{1}}+{{n}_{2}}}$
So, the answer is option D.

Note: The important equations used to solve this problem are $\log (AB)=\log A+\log B$ and $\log ({{A}^{n}})=n\log (A)$. Also while finding the G, we sometimes tend to make mistakes such as interchange ${{G}_{1}}$ and ${{G}_{2}}$. In the competitive exam, be careful while marking the answers because in the given option itself the option (C) and option (D) seems to be similar.