Question

The least value of 'a' for which $5^{1+x}+5^{1-x},\frac{a}{2},25^x+25^{-x}$ are three consecutive terms of an A.P. is

• A. 10
• B. 5
• C. 12
• D. none of these.

Answer 1

Answer: (C)

12

Answer 2

Since $5^{1+x} +5^{1-x}, \frac{a}{2}, \left(25\right)^{x} +\left(25\right)^{-x}$ are in $A.P.$ $\therefore a= 5^{1+x}+5^{1-x}+25^{x}+25^{-x}$ $=5.5^{x}+5.5^{-x} +5^{2x}+5^{-2x}$ $= 5t+\frac{5}{t}+t^{2}+\frac{1}{t^{2}}$ where $5^{x}= t$ $=\left(t-\frac{1}{t}\right)^{2} +5\left(\sqrt{t} -\frac{1}{\sqrt{t}}\right)^{2} +12 \ge 12$ $\therefore a \ge 12$ $\therefore$ least value of $a=12.$