JEE Mains · Maths · STD 11 - 8. sequence and series
For \(\mathrm{x} \geq 0\), the least value of \(\mathrm{K}\), for which \(4^{1+\mathrm{x}}+4^{1-\mathrm{x}}\), \(\frac{\mathrm{K}}{2}, 16^{\mathrm{x}}+16^{-\mathrm{x}}\) are three consecutive terms of an \(A.P.\) is equal to :
- A \(10\)
- B \(4\)
- C \(8\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\( \mathrm{k}=4\left(4^{\mathrm{x}}+\frac{1}{4^{\mathrm{x}}}\right)+\left(4^{2 \mathrm{x}}+\frac{1}{4^{2 \mathrm{x}}}\right) \) \( \quad \geq 2 \quad \geq 2 \) \( \mathrm{k} \geq 10\)
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