Question

If $x \in R$ the numbers ${5^{1 + x}} + {5^{1 - x}},\dfrac{a}{2},{25^x} + {25^{ - x}}$ form an A.P. then ‘a’ must lie in the interval
A.$\left[ {1,5} \right]$
B.$\left[ {2,5} \right]$
C.$\left[ {5,12} \right]$
D.$[12,\infty )$

Answer 1

Hint: At first we will form an equation of the three numbers ${5^{1 + x}} + {5^{1 - x}},\dfrac{a}{2},{25^x} + {25^{ - x}}$, as they are in A.P. using the property of the A.P. that the sum of the first and third numbers is equal to the twice the second number.
Then we will solve that equation for a, to find the range of the a, using the property that A.M. of any two numbers is always greater or equal to the G.M. of the two numbers.

Complete step-by-step answer:
Given data: ${5^{1 + x}} + {5^{1 - x}},\dfrac{a}{2},{25^x} + {25^{ - x}}$ form an A.P.
We know that if there are three numbers in A.P. then the sum of the first and third numbers is equal to the twice the second number.
Now since the numbers ${5^{1 + x}} + {5^{1 - x}},\dfrac{a}{2},{25^x} + {25^{ - x}}$are in A.P.
\[ \Rightarrow a = \left( {{5^{1 + x}} + {5^{1 - x}}} \right) + \left( {{{25}^x} + {{25}^{ - x}}} \right)\]
We know that ${a^{b + c}} = {a^b}{a^c}$, using this property
\[ \Rightarrow a = \left( {5 \times {5^x} + 5 \times {5^{ - x}}} \right) + \left( {{{25}^x} + {{25}^{ - x}}} \right)\]
Taking 5 commons from the first bracket
\[ \Rightarrow a = 5\left( {{5^x} + {5^{ - x}}} \right) + \left( {{{25}^x} + {{25}^{ - x}}} \right)................(i)\]
Now if we take two numbers let say $'m'$ and it's reciprocal $'\dfrac{1}{m}'$
We know that A.M. of any two numbers is always greater or equal to the G.M. of the two numbers
i.e. $\dfrac{{m + \dfrac{1}{m}}}{2} \geqslant \sqrt {m \times \dfrac{1}{m}} $
On cross multiplication and simplification we get,
$ \Rightarrow m + \dfrac{1}{m} \geqslant 2$
Therefore the range of the expression, $m + \dfrac{1}{m}$ is $[2,\infty )$
Now we can say that the range of the expression \[5\left( {{5^x} + {5^{ - x}}} \right) + \left( {{{25}^x} + {{25}^{ - x}}} \right)\] will be \[[5\left( 2 \right) + \left( 2 \right),\infty )\]
i.e. \[[12,\infty )\]
And from equation(i),
\[a = 5\left( {{5^x} + {5^{ - x}}} \right) + \left( {{{25}^x} + {{25}^{ - x}}} \right)\]
Therefore, \[a \in [12,\infty )\]
Hence, Option (D) is correct


Note: In the above solution we found the range set of the sum of a number and its reciprocal, an alternative method for the same can be
Let us suppose $m + \dfrac{1}{m} = t$
On simplifying by taking LCM we get,
$ \Rightarrow {m^2} + 1 = tm$
$ \Rightarrow {m^2} - tm + 1 = 0$
Now we know that, $m = \dfrac{{t \pm \sqrt {{t^2} - 4} }}{2}$
From this, we can say that ${t^2} - 4 \geqslant 0$, since square root term cannot be negative
$ \Rightarrow {t^2} \geqslant 4$
$ \Rightarrow t \in ( - \infty , - 2] \cup [2,\infty )$
But according to the values required, m is either ${25^x}$ or ${5^x}$i.e. the exponential of the positive constant which can never be negative neither their sum.
Which means,
Substituting the value of t,
$ \Rightarrow m + \dfrac{1}{m} \in [2,\infty )$, it is the same range set as we acquired in the above solution.