Answer
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Hint: Here, we will first find the arithmetic mean and geometric mean of the terms in the expression. Then, we will use the relation between arithmetic mean and geometric mean to form an inequation. Finally, we will use the given information to find the minimum value of the expression \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\].
Formula Used:
We will use the following formulas:
The arithmetic mean of the \[n\] numbers \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] is given by the formula \[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + {a_n}}}{n}\].
The geometric mean of the \[n\] numbers \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] is given by the formula \[G.M. = \sqrt[n]{{{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}}}\].
Complete step-by-step answer:
We will use the formula for A.M. and G.M. to find the minimum value of \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\].
The number of terms in the sum \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is \[n\].
Therefore, using the formula \[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + {a_n}}}{n}\], we get the arithmetic mean as
\[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n}\]
The number of terms in the sum \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is \[n\].
Therefore, using the formula \[G.M. = \sqrt[n]{{{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}}}\], we get the geometric mean as
\[G.M. = {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}2{a_n}} \right)^{1/n}}\]
Now, we know that the arithmetic mean is always greater than or equal to the geometric mean.
Therefore, we get
\[ \Rightarrow A.M. \ge G.M.\]
Substituting \[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n}\] and \[G.M. = {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}2{a_n}} \right)^{1/n}}\] in the inequation, we get
\[ \Rightarrow \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}2{a_n}} \right)^{1/n}}\]
Rewriting the inequation, we get
\[ \Rightarrow \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}} \right)^{1/n}}\]
It is given that the number \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] are positive real numbers whose product is a fixed number \[c\].
Therefore, we get
\[{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n} = c\]
Substituting \[{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n} = c\] in the inequation \[\dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}} \right)^{1/n}}\], we get
\[ \Rightarrow \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2c} \right)^{1/n}}\]
Multiplying both sides by \[n\], we get
\[ \Rightarrow n\left( {\dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n}} \right) \ge n{\left( {2c} \right)^{1/n}}\]
Thus, we get
\[ \Rightarrow {a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n} \ge n{\left( {2c} \right)^{1/n}}\]
Therefore, the value of the expression \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is greater than or equal to \[n{\left( {2c} \right)^{1/n}}\].
Thus, the minimum value of the expression \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is \[n{\left( {2c} \right)^{1/n}}\].
The correct option is option (a).
Note: We multiplied both sides of the inequation \[\dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2c} \right)^{1/n}}\] by \[n\]. Since the number of terms cannot be negative, \[n\] is a positive integer. Therefore, we could multiply both sides of the inequation by \[n\] without changing the sign of the inequation.
Here we used geometric mean and arithmetic mean to solve the question. These are the two types of mean and the third type of mean is harmonic mean.
Formula Used:
We will use the following formulas:
The arithmetic mean of the \[n\] numbers \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] is given by the formula \[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + {a_n}}}{n}\].
The geometric mean of the \[n\] numbers \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] is given by the formula \[G.M. = \sqrt[n]{{{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}}}\].
Complete step-by-step answer:
We will use the formula for A.M. and G.M. to find the minimum value of \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\].
The number of terms in the sum \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is \[n\].
Therefore, using the formula \[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + {a_n}}}{n}\], we get the arithmetic mean as
\[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n}\]
The number of terms in the sum \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is \[n\].
Therefore, using the formula \[G.M. = \sqrt[n]{{{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}}}\], we get the geometric mean as
\[G.M. = {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}2{a_n}} \right)^{1/n}}\]
Now, we know that the arithmetic mean is always greater than or equal to the geometric mean.
Therefore, we get
\[ \Rightarrow A.M. \ge G.M.\]
Substituting \[A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n}\] and \[G.M. = {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}2{a_n}} \right)^{1/n}}\] in the inequation, we get
\[ \Rightarrow \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}2{a_n}} \right)^{1/n}}\]
Rewriting the inequation, we get
\[ \Rightarrow \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}} \right)^{1/n}}\]
It is given that the number \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] are positive real numbers whose product is a fixed number \[c\].
Therefore, we get
\[{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n} = c\]
Substituting \[{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n} = c\] in the inequation \[\dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}} \right)^{1/n}}\], we get
\[ \Rightarrow \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2c} \right)^{1/n}}\]
Multiplying both sides by \[n\], we get
\[ \Rightarrow n\left( {\dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n}} \right) \ge n{\left( {2c} \right)^{1/n}}\]
Thus, we get
\[ \Rightarrow {a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n} \ge n{\left( {2c} \right)^{1/n}}\]
Therefore, the value of the expression \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is greater than or equal to \[n{\left( {2c} \right)^{1/n}}\].
Thus, the minimum value of the expression \[{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}\] is \[n{\left( {2c} \right)^{1/n}}\].
The correct option is option (a).
Note: We multiplied both sides of the inequation \[\dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + 2{a_n}}}{n} \ge {\left( {2c} \right)^{1/n}}\] by \[n\]. Since the number of terms cannot be negative, \[n\] is a positive integer. Therefore, we could multiply both sides of the inequation by \[n\] without changing the sign of the inequation.
Here we used geometric mean and arithmetic mean to solve the question. These are the two types of mean and the third type of mean is harmonic mean.
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