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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Sequences and Series

NDA 2021 · Mathematics · Sequences and Series

If x, y, z are in GP, then which of the following is/are correct?...

If x,y,zx, y, z are in GP, then which of the following is/are correct? 1. ln⁡(3x),ln⁡(3y),ln⁡(3z)\ln (3 x), \ln (3 y), \ln (3 z) are in AP 2. xyz+ln⁡(),xyz+ln⁡(y),xyz+ln⁡(z)x y z+\ln (), x y z+\ln (y), x y z+\ln (z) are in HP Select the correct answer using the code given below.

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

Since, x,yandzx, y and z are in G.P ∴y2=xz\therefore y^2=x z Taking log⁡\log at base ee both sides ln⁡(y2)=ln⁡(xz)⇒2ln⁡(y)=ln⁡(x)+ln⁡(z)\ln \left(y^2\right)=\ln (x z)\Rightarrow 2 \ln (y)=\ln (x)+\ln (z) ....(i) ∴ln⁡(x),ln⁡(y) and  ln⁡(z) are in A.P. adding xyzSo, xyz+ln⁡(x),xyz+ln⁡(y),xyz+ln⁡(z) are also in A.P. Now, ln⁡(3x)+ln⁡(3z)=ln⁡(3)+ln⁡(x)+ln⁡(3)+ln⁡(z)=2ln⁡(3)+ln⁡(x)+ln⁡(z)\begin{aligned} & \therefore \ln (x), \ln (y) \text{\text{ and } } \ln (z) \text{ are in A.P. } \\ & \text{adding } x y z \\ & \text{So, } x y z+\ln (x), x y z+\ln (y), x y z+\ln (z) \text{ are also in A.P. } \\ & \text{Now, } \ln (3 x)+\ln (3 z)=\ln (3)+\ln (x)+\ln (3)+\ln (z) \\ & =2 \ln (3)+\ln (x)+\ln (z)\end{aligned}=2 \ln (3)+2 \ln (y) \quad[fromfrom(\mathrm{i})]=2ln⁡(3y)∴ln⁡(3x),ln⁡(3y) and  ln⁡(3z) are in A.P. \begin{aligned} & =2 \ln (3 y) \\ & \therefore \ln (3 x), \ln (3 y) \text{\text{ and } } \ln (3 z) \text{ are in A.P. }\end{aligned}$
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