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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Sequences and Series

NDA 2024 · Mathematics · Sequences and Series

Let x > 1, y > 1, z > 1 be in GP. Then (1)/(1+ x), (1)/(1+ y),...

Let x>1,y>1,z>1x > 1, y > 1, z > 1 be in GP. Then 11+ln⁡x,11+ln⁡y,11+ln⁡z\frac{1}{1+\ln x}, \frac{1}{1+\ln y}, \frac{1}{1+\ln z} are

Options

  1. Ain AP
  2. Bin GP
  3. Cin HP
  4. Dneither in AP nor in GP nor in HP

Correct answer

C. in HP

Step-by-step solution

Since, x,y,zare\mathrm{x}, \mathrm{y}, \mathrm{z} are in G.P.⇒y2=xz⇒2ln⁡y=ln⁡x+ln⁡z⇒ln⁡y−ln⁡x=ln⁡z−ln⁡y⇒ln⁡x,ln⁡y,ln⁡zare⁡in⁡AP.⇒(1+ln⁡x),(1+ln⁡y),(1+ln⁡z)are in AP⇒1(1+ln⁡x),1(1+ln⁡y),1(1+ln⁡z) are in AP.\begin{aligned} & \Rightarrow y^2=x z \Rightarrow 2\ln y=\ln x+\ln z \\ & \Rightarrow \ln y-\ln x=\ln z-\ln y \\ & \Rightarrow \ln x, \ln y, \ln z \operatorname{are} \operatorname{in} A P . \\ & \Rightarrow(1+\ln x),(1+\ln y),(1+\ln z) \text{are in } A P \\ & \Rightarrow\frac{1}{(1+\ln x)}, \frac{1}{(1+\ln y)}, \frac{1}{(1+\ln z)} \text{ are in } A P . \end{aligned}
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