For x geq 0,the least value of K,for which 4^ | vedclass

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(A) Let the three terms be $a = 4^{1+x} + 4^{1-x}$,$b = \frac{K}{2}$,and $c = 16^x + 16^{-x}$.Since they are in $A.P.$,$2b = a + c$.Substituting the v

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$10$

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(A) Let the three terms be $a = 4^{1+x} + 4^{1-x}$,$b = \frac{K}{2}$,and $c = 16^x + 16^{-x}$.Since they are in $A.P.$,$2b = a + c$.Substituting the values,$2(\frac{K}{2}) = 4(4^x + 4^{-x}) + (4^{2x} + 4^{-2x})$.Let $y = 4^x + 4^{-x}$. Since $x \geq 0$,$y \geq 2$ by the $AM-GM$ inequality.Then $4^{2x} + 4^{-2x} = (4^x + 4^{-x})^2 - 2 = y^2 - 2$.So,$K = 4y + y^2 - 2 = y^2 + 4y - 2$.To find the least value of $K$ for $y \geq 2$,we evaluate $f(y) = y^2 + 4y - 2$ at $y = 2$.$f(2) = 2^2 + 4(2) - 2 = 4 + 8 - 2 = 10$.Thus,the least value of $K$ is $10$.

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

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