For x ⩾ 0 , the least value of K , for which 4 1 + x + 4 1 − x , 2 K , 1 6 x + 1 6 − x are three consecutive terms of an A.P., is equal to :

k = 4 ( 4 x + 4 x 1 ) + ( 4 2 x + 4 2 x 1 ) ≥ 2 ≥ 2 k ≥ 10

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesA.P.Easy2 minPYQ_2024

MathematicsEasystatement

For $x ⩾ 0$ , the least value of $K$ , for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 1 6^{x} + 1 6^{- x}$ are three consecutive terms of an A.P., is equal to :

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Answer:

Solution:

$k = 4 (4^{x} + \frac{1}{4 ^{x}}) + (4^{2 x} + \frac{1}{4 ^{2 x}}) \geq 2 \geq 2$ $k \geq 10$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:A.P.

⏱ 2mℹ️ Source: PYQ_2024

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