Question

If \[x\in R\] and the numbers \[{{5}^{\left( 1-x \right)}}+{{5}^{\left( x+1 \right)}},\dfrac{a}{2},\left( {{25}^{x}}+{{25}^{-x}} \right)\] form an A.P., then a must lie in the interval:
A. \[\left[ 12,\infty \right]\]
B. \[\left[ -5,5 \right]\]
C. \[\left[ -12,12 \right]\]
D. \[\left[ 8,\infty \right]\]

Answer 1

Hint: In the above question, we will use the concept that if three numbers a, b and c are in A.P. then \[b=\dfrac{a+c}{2}\]. Also, we will use the concept that if we have positive numbers then it follows the inequality as shown as follows:
\[A.M.\ge G.M.\] where A.M. is arithmetic mean and G.M. is geometric mean.

Complete step-by-step answer:
We have been given the numbers \[{{5}^{\left( 1-x \right)}}+{{5}^{\left( x+1 \right)}},\dfrac{a}{2},\left( {{25}^{x}}+{{25}^{-x}} \right)\] form an A.P.
We know that if a, b and c are in A.P. then \[b=\dfrac{a+c}{2}\].
\[\Rightarrow \dfrac{a}{2}=\dfrac{{{5}^{\left( 1-x \right)}}+{{5}^{\left( x+1 \right)}}+\left( {{25}^{x}}+{{25}^{-x}} \right)}{2}\]
On cross multiplying the equations by 2 we get,
\[\begin{align}
  & \Rightarrow 2\times \dfrac{a}{2}={{5}^{\left( 1-x \right)}}+{{5}^{\left( x+1 \right)}}+\left( {{25}^{x}}+{{25}^{-x}} \right) \\
 & \Rightarrow a=\dfrac{5}{{{5}^{x}}}+5\times {{5}^{x}}+{{\left( {{5}^{2}} \right)}^{x}}+{{\left( {{5}^{2}} \right)}^{-x}} \\
 & \Rightarrow a=\dfrac{5}{{{5}^{x}}}+5\times {{5}^{x}}+{{5}^{2x}}+{{5}^{-2x}} \\
 & \Rightarrow a=\dfrac{5}{{{5}^{x}}}+5\times {{5}^{x}}+{{5}^{2x}}+\dfrac{1}{{{\left( {{5}^{2}} \right)}^{x}}} \\
\end{align}\]
Let us suppose that \[{{5}^{x}}=t\] where t > 0 since \[{{5}^{x}}\] > 0 for all real values of x.
\[\begin{align}
  & \Rightarrow a=\dfrac{5}{t}+5t+{{\left( t \right)}^{2}}+\dfrac{1}{{{\left( t \right)}^{2}}} \\
 & \Rightarrow a=5\left( \dfrac{1}{t}+t \right)+\left( {{t}^{2}}+\dfrac{1}{{{t}^{2}}} \right) \\
\end{align}\]
We know that if we have positive numbers then they follow an inequality as follows:
\[A.M.\ge G.M.\], where A.M. is arithmetic mean and G.M. is geometric mean.
For \[\dfrac{1}{t},t\] we have \[A.M.\ge G.M.\]
Since the A.M. and G.M. of two numbers a and b are \[\dfrac{a+b}{2}\] and \[\sqrt{ab}\] respectively.
\[\begin{align}
  & \Rightarrow \dfrac{t+\dfrac{1}{t}}{2}\ge \sqrt{t\times \dfrac{1}{t}} \\
 & \Rightarrow t+\dfrac{1}{t}\ge 2\sqrt{1} \\
 & \Rightarrow t+\dfrac{1}{t}\ge 2 \\
\end{align}\]
For \[\dfrac{1}{{{t}^{2}}},{{t}^{2}}\] we have \[A.M.\ge G.M.\]
Since the A.M. and G.M. of two numbers a and b are \[\dfrac{a+b}{2}\] and \[\sqrt{ab}\] respectively.

\[\begin{align}
  & \Rightarrow \dfrac{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}}{2}\ge \sqrt{{{t}^{2}}\times \dfrac{1}{{{t}^{2}}}} \\
 & \Rightarrow {{t}^{2}}+\dfrac{1}{{{t}^{2}}}\ge 2\sqrt{1} \\
 & \Rightarrow {{t}^{2}}+\dfrac{1}{{{t}^{2}}}\ge 2 \\
\end{align}\]
On substituting the minimum values of \[\left( t+\dfrac{1}{t} \right)\] and \[\left( {{t}^{2}}+\dfrac{1}{{{t}^{2}}} \right)\] we get as follows:
\[\begin{align}
  & \Rightarrow a=5(2)+2 \\
 & \Rightarrow a=12 \\
\end{align}\]
Hence the minimum value of ‘a’ is 12 and its maximum value is equal to infinity as \[\left( t+\dfrac{1}{t} \right)\] and \[\left( {{t}^{2}}+\dfrac{1}{{{t}^{2}}} \right)\] tends to infinity.
Therefore, the correct answer of the above question is option A.

Note: Be careful while calculating as there is a chance of sign mistake during simplifying the equation.
Also, remember that for any positive numbers \[A.M.\ge G.M.\ge H.M.\] The transformation of the terms in the first part of the solution must be done carefully and properly. Any mistake in formulas used can lead to a completely different answer.