For $x \geqslant 0$, the least value of $\mathrm{K}$, for which $4^{1+x}+4^{1-x}, \frac{\mathrm{K}}{2}, 16^x+16^{-x}$ are three consecutive terms of an A.P., is equal to :

Question

Aicharya · Accepted Answer

$\begin{gathered}\mathrm{k}=4\left(4^{\mathrm{x}}+\frac{1}{4^{\mathrm{x}}}\right)+\left(4^{2 \mathrm{x}}+\frac{1}{4^{2 \mathrm{x}}}\right) \ \qquad \geq 2 \quad\qquad\qquad \geq 2\end{gathered}$ $\mathrm{k} \geq 10$

For $x \geqslant 0$, the least value of $\mathrm{K}$, for which $4^{1+x}+4^{1-x}, \frac{\mathrm{K}}{2},…

Solution