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: [2024]

Question

For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

Smart Achievers · Accepted Answer

(2)

We have, $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are in A.P.

$\Rightarrow 2 \frac{K}{2} = (4^{1 + x} + 4^{1 - x}) + (16^{x} + 16^{- x})$

$\Rightarrow K = 4 \cdot 4^{x} + \frac{4}{4^{x}} + 4^{2 x} + \frac{1}{4^{2 x}}$

$= 4 (4^{x} + \frac{1}{4^{x}}) + (4^{2 x} + \frac{1}{4^{2 x}})$ $\geq 4 \cdot 2 + 2$ $[∵ A . M . \geq G . M .]$

$= 10 \Rightarrow K \geq 10$

So, least value of $K$ is 10

Solution

Q.
For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]

1 4

2 10

3 8

4 16

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Solution

Q. For x≥0, the least value of K, for which 41+x+41-x, K2, 16x+16-x are three consecutive terms of an A.P., is equal to: [2024]

1 4

2 10

3 8

4 16

Ans. (2) We have, 41+x+41-x,K2,16x+16-x are in A.P. ⇒2K2=(41+x+41-x)+(16x+16-x) ⇒K=4·4x+44x+42x+142x =4(4x+14x)+(42x+142x) ≥4·2+2 [∵A.M.≥G.M.] =10⇒K≥10 So, least value of K is 10

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Q.
For $x \geq 0,$ the least value of $K,$ for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 16^{x} + 16^{- x}$ are three consecutive terms of an A.P., is equal to: [2024]