Question

# For what values of the parameter a are there values of x such that  $$\displaystyle 5^{1+x}+5^{1-x}, a/2, 25^{x}+25^{-x}$$ are three consecutive terms of an A.P.?

Solution

## Since $$\displaystyle 5^{1+x}+5^{1-x}, a/2, 25^{x}+25^{-x}$$ are in A. P., we have$$\displaystyle 2a/2= 5^{1+x}+5^{1-x}+25^{x}+25^{-x},$$Now put $$\displaystyle 5^{x}= t$$ so that $$\displaystyle t> 0,$$ we then have$$\displaystyle a= 5t+5/t^{2}+1/t^{2}= +ive$$ ...(1)$$\displaystyle = \left ( t^{2}+1/t^{2} \right )+5\left ( t+1/t \right )\geq 2+5.2= 12$$because $$\displaystyle A.M.\geq G.M.$$Also when $$\displaystyle a\geq 12$$ then (1) becomes$$\displaystyle a= \left ( t+\frac{1}{t} \right )^{2}-2+5\left ( t+\frac{1}{t} \right )$$$$\Rightarrow\displaystyle y^{2}+5y-\left ( a+2 \right )= 0$$where $$\displaystyle y= t+\frac{1}{t}\geq 2\:as\:A.M.\geq G.M.$$$$\displaystyle \therefore y= \frac{-5\pm \sqrt{25+4a+8}}{2}$$$$\displaystyle y= \frac{-5\pm \sqrt{33+4a}}{2}\geq \frac{-5\pm \sqrt{33+48}}{2}$$$$\displaystyle = \frac{-5\pm 9}{2}= \frac{9-5}{2}= 2 \therefore y\geq 2$$$$\displaystyle \therefore y= t+\frac{1}{t}= 5^{x}+5^{-x}\geq 2$$which we know is true and will give a solution for $$x.$$Maths

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