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Question: If \[{x_i} > 0\], \[i = 1,2,3,....,50\] and \[{x_1} + {x_2} + ... + {x_{50}} = 50\] then the minimum...

If xi>0{x_i} > 0, i=1,2,3,....,50i = 1,2,3,....,50 and x1+x2+...+x50=50{x_1} + {x_2} + ... + {x_{50}} = 50 then the minimum value of 1x1+1x2+...+1x50\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + ... + \dfrac{1}{{{x_{50}}}} equals to
A). 5050
B). (50)2{\left( {50} \right)^2}
C). (50)3{\left( {50} \right)^3}
D). (50)4{\left( {50} \right)^4}

Explanation

Solution

In the given question, we have been given the sum of an arithmetic progression. We have to calculate the value of the minimum sum of the harmonic progression of this arithmetic progression. We are going to solve this by using the relation between the arithmetic mean and the harmonic mean.

Complete step by step solution:
We know, AM≥HMAM \ge HM, so
x1+x2+...+x5050≥501x1+1x2+...+1x50\dfrac{{{x_1} + {x_2} + ... + {x_{50}}}}{{50}} \ge \dfrac{{50}}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + ... + \dfrac{1}{{{x_{50}}}}}}
Now, it has been given that, x1+x2+...+x50=50{x_1} + {x_2} + ... + {x_{50}} = 50, so,
5050≥501x1+1x2+...+1x50\dfrac{{50}}{{50}} \ge \dfrac{{50}}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + ... + \dfrac{1}{{{x_{50}}}}}}
Hence, 1≥501x1+1x2+...+1x501 \ge \dfrac{{50}}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + ... + \dfrac{1}{{{x_{50}}}}}}
or 1x1+1x2+...+1x50≥50\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + ... + \dfrac{1}{{{x_{50}}}} \ge 50
Hence, the correct option is A.

Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we write the formula which connects the two things. We only need to pay attention to the formula of the relation between the means of the two different entities. It is the only thing around which the answer revolves.