Question

If ${{a}_{1}},{{a}_{2}},....{{a}_{n}}$ are positive numbers such that ${{a}_{1}}\cdot {{a}_{2}}\cdot {{a}_{3}}.....{{a}_{n}}=1$ then their sum is
$\left( A \right)$ a positive integer
$\left( B \right)$ divisible by $n$
$\left( C \right)$ never less than $n$
$\left( D \right)$ none of these

Answer 1

Hint: In this question we have been given a series of terms for which we have been given that all the numbers in the series are positive. We know that the product of all the terms in the series is equal to $1$. We will solve this question by using the mathematical property that the arithmetic mean is greater than or equal to the geometric mean for the same series. We will then substitute the values given in the series and simplify to get the required solution.

Complete step-by-step solution:
We know that ${{a}_{1}},{{a}_{2}},....{{a}_{n}}$ are positive numbers.
We also know that ${{a}_{1}}\cdot {{a}_{2}}\cdot {{a}_{3}}.....{{a}_{n}}=1$.
Now we know that for the same series the arithmetic mean is greater than or equal to the geometric mean. Therefore, mathematically we can write it as:
$\Rightarrow A.M\ge G.M$
Now since there are total $n$ number of terms in the series, we can write the arithmetic mean as:
$\Rightarrow A.M=\left( \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} \right)$
And the geometric mean of the terms can be written as:
$\Rightarrow G.M={{\left( {{a}_{1}}\cdot {{a}_{2}}\cdot {{a}_{3}}.....{{a}_{n}} \right)}^{\dfrac{1}{n}}}$
On equating both, we get:
$\Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} \right)\ge {{\left( {{a}_{1}}\cdot {{a}_{2}}\cdot {{a}_{3}}.....{{a}_{n}} \right)}^{\dfrac{1}{n}}}$
Now we know that ${{a}_{1}}\cdot {{a}_{2}}\cdot {{a}_{3}}.....{{a}_{n}}=1$ therefore, on substituting, we get:
$\Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} \right)\ge {{\left( 1 \right)}^{\dfrac{1}{n}}}$
Now we know that $1$ raised to anything is $1$ therefore, we can write:
$\Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} \right)\ge 1$
On transferring the term $n$ from the left-hand side to the right-hand side, we get:
$\Rightarrow \left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}} \right)\ge n$
Therefore, from the above expression, we can say that the sum of the terms in the series will never be less than $n$.
Hence the correct answer is option $\left( C \right)$.

Note: In this question we have used the relation between the arithmetic mean and geometric mean for the sequence of terms given. There also exists another type of mean which is the harmonic mean which could be written for the above terms as $H.M=\dfrac{n}{\left( \dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{3}}}+....+\dfrac{1}{{{a}_{n}}} \right)}$. The relationship between all three is that $A.M\ge G.M\ge H.M$.