Question

The minimum value of the sum of real numbers ${a^{ - 5}},{a^{ - 4}},3{a^{ - 3}},1,{a^8},{a^{10}}$ with $a > 0$ is
1. $9$
2. $8$
3. $2$
4. $1$

Answer 1

We must know the meaning of the Arithmetic mean and the Geometric Mean and the difference between them. We will find the Arithmetic mean and the Geometric Mean using the respective formulas as $AM = \dfrac{{{a_1},{a_2},...,{a_n}}}{n}$ and $GM = {\left( {{a_1},{a_2},...,{a_n}} \right)^{\dfrac{1}{n}}}$ respectively. Use the relation between the Arithmetic mean and the Geometric Mean that is $AM \geqslant GM$. Simplify it as much as possible. You will get the required value.

Complete step-by-step solution:
Arithmetic means a number that is obtained by dividing the sum of the elements of a set by the number of values in the set.
Consider, if ${a_1},{a_2},...,{a_n}$ are the observations, then the A.M. is defined as
$AM = \dfrac{{{a_1},{a_2},...,{a_n}}}{n}$
The Geometric Mean (GM) is the average value or mean which signifies the central tendency of the set of numbers by finding the product of their values.
In other words, the geometric mean is defined as the nth root of the product of n numbers
Consider, if ${a_1},{a_2},...,{a_n}$ are the observations, then the G.M is defined as
$GM = {\left( {{a_1},{a_2},...,{a_n}} \right)^{\dfrac{1}{n}}}$
$AM \geqslant GM$
$\dfrac{{\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}}}}{8} \geqslant {\left( {\dfrac{1}{{{a^5}}} \times \dfrac{1}{{{a^4}}} \times \dfrac{1}{{{a^3}}} \times \dfrac{1}{{{a^3}}} \times \dfrac{1}{{{a^3}}} \times 1 \times {a^8} \times {a^{10}}} \right)^{\dfrac{1}{8}}}$
On simplifying this we get ,
$\dfrac{{\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}}}}{8} \geqslant {\left( 1 \right)^{\dfrac{1}{8}}}$
Hence we get ,
$\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}} \geqslant 8{\left( 1 \right)^{\dfrac{1}{8}}}$
Therefore we have $\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}} \geqslant 8$
Therefore The minimum value of the sum of real numbers ${a^{ - 5}},{a^{ - 4}},3{a^{ - 3}},1,{a^8},{a^{10}}$ is $8$.
Therefore option (2) is the correct answer.

Note: Arithmetic mean represents a number that is obtained by dividing the sum of the elements of a set by the number of values in the set. The Geometric Mean (GM) is the average value or mean which signifies the central tendency of the set of numbers by finding the product of their values. The geometric mean is different from the arithmetic mean.