If x ∈ R , the numbers 5 1 + x + 5 1 − x , 2 a , 2 5 x + 2 5 − x form an A.P., then a must lie in the interval [0pt]

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MathematicsSequence & SeriesMeans / inequalityMedium2 minQB

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If $x \in R$ , the numbers $5^{1 + x} + 5^{1 - x}, \frac{a}{2}, 2 5^{x} + 2 5^{- x}$ form an A.P., then $a$ must lie in the interval [0pt]

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$\frac{a}{2} = \frac{( 5 ^{1 + x} + 5 ^{1 - x} ) + ( 2 5 ^{x} + 2 5 ^{- x} )}{2} \Rightarrow a = 5 (5^{x} + 5^{- x}) + (5^{2 x} + 5^{- 2 x})$ . Minimum of $5^{x} + 5^{- x}$ is 2 and of $5^{2 x} + 5^{- 2 x}$ is 2, both at $x = 0$ . So $a_{min} = 5 \times 2 + 2 = 12$ , and $a \to \infty$ as $∣ x ∣ \to \infty$ . Hence $a \in [12, \infty)$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Means / inequality

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