For x >= 0, the least value of K, for which 4^1 + x + 4^1 - x,(K / 2),16^x + 16^- x are three consecutive terms of an A.P., is equal to

(b) : We have, 4^1+x+4^1-x, (K / 2), 16^x+16^-x are in A.P. => 2 (K / 2)=(4^1+x+4^1-x)+(16^x+16^-x) \ => K=4 * 4^x+(4 / 4^x)+4^2 x+frac14^2 x \ =4(4^x+(1 / 4^x))+(4^2 x+frac14^2 x) \ >= 4 * 2+2[Because A.M. >= G.M. ] \ =10 => K >= 10 \ So, least value of K is mathbf1 0.

For x ? 0, the least value of K, for which 4^{1 + x} +... | Sequence & Series (Mathematics)

(b) : We have, $4^{1+x}+4^{1-x}, \frac{K}{2}, 16^x+16^{-x}$ are in A.P. $\begin{aligned} & \Rightarrow 2 \frac{K}{2}=\left(4^{1+x}+4^{1-x}\right)+\left(16^x+16^{-x}\right) \\ & \Rightarrow K=4 \cdot 4^x+\frac{4}{4^x}+4^{2 x}+\frac{1}{4^{2 x}} \\ & =4\left(4^x+\frac{1}{4^x}\right)+\left(4^{2 x}+\frac{1}{4^{2 x}}\right) \\ & \geq 4 \cdot 2+2[\because \text { A.M. } \geq \text { G.M. }] \\ & =10 \Rightarrow K \geq 10 \end{aligned}$ So, least value of $K$ is $\mathbf{1 0}$ .

Explanation

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