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Assertion (A):If for any real $x,$ $2^{1 + x} + 2^{1 - x}, λ, 3^{x} + 3^{- x}$ are three equidistant terms of an A.P then $\sin^{- 1} (\frac{λ}{3}) = \frac{π}{2}$
Reason(R): $A + B \geq 2 \sqrt{A B}$

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Detailed Solution

$2 λ = 2^{1 + x} + 2^{1 - x} + 3^{x} + 3^{- x}$

$\geq 2 \sqrt{2^{1 + x} {.2}^{1 - x}} + 2 \sqrt{3^{x} {.3}^{- x}}$

$\geq 2 (2) + 2 (1) & -1 \leq \frac{λ}{3} \leq 1$

$\geq 6 -3 \leq λ \leq 3$

$∴ λ \geq 3$

$∴ λ = 3$

$∴ \sin^{- 1} (\frac{λ}{3}) = \frac{π}{2}$

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