Question

If \[{a_1},{a_2}..............{a_n}\]are positive real numbers whose product is a fixed number c, then the minimum value of \[{a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n}\;\] is
A) \[n{(2c)^{(1/n)}}\]
B) \[(n + 1){c^{(1/n)}}\]
C) \[2n{c^{(1/n)}}\]
D) \[(n + 1){(2c)^{(1/n)}}\]

Answer 1

Hint: In this question we have to find the sum of given number of terms. First find arithmetic mean and geometric mean of given terms then use concept that Arithmetic mean is always greater than or equal to geometric mean.

Formula used: Arithmetic mean is given as
\[\dfrac{{{a_1} + {a_2} + {a_3}...{a_n}}}{n}\]
Where
\[{a_1},{a_2},{a_3},...{a_n}\] Any given real number
Geometric mean is given as
\[{({a_1}.{a_2}.{a_3}....{a_n})^{\dfrac{1}{n}}}\]
\[{a_1},{a_2},{a_3},...{a_n}\] Any given real number

Complete step by step solution: Given: \[{a_1},{a_2}..............{a_n}\] are positive real numbers whose product is a fixed number c
We know that arithmetic mean is greater than or equal to geometric mean
\[\dfrac{{({a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n})\;}}{n} \ge {({a_1}.{a_2}.{a_3}....2{a_n})^{\dfrac{1}{n}}}\]
\[{a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n} \ge n{(2c)^{\dfrac{1}{n}}}\]
Now minimum value of \[{a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n}\;\]is given as
\[{a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n} = n{(2c)^{\dfrac{1}{n}}}\]

Thus, Option (A) is correct.

Note: We must remember that arithmetic mean of given number is always greater than or equal to geometric mean of same number.
Don’t try to find individual term, multiplication of each term in order to equate it with c.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.