Q155 / 270 · NDA 2025 (Phase 2) · Mathematics
If $(10 + \log_{10} x)$, $(10 + \log_{10} y)$ and $(10 + \log_{10} z)$ are in AP, then consider the following statements:
I. The GM of $x$ and $z$ is $y^2$.
II. The AM of $\log_{10} x$ and $\log_{10} z$ is $\log_{10} y$.
Which of the statements given above is/are correct?
I. The GM of $x$ and $z$ is $y^2$.
II. The AM of $\log_{10} x$ and $\log_{10} z$ is $\log_{10} y$.
Which of the statements given above is/are correct?
- AI only
- Correct answerBII only
- CBoth I and II
- DNeither I nor II
Solution
Given $(10 + \log_{10} x)$, $(10 + \log_{10} y)$, and $(10 + \log_{10} z)$ are in AP.
$2(10 + \log_{10} y) = (10 + \log_{10} x) + (10 + \log_{10} z)$
$20 + 2\log_{10} y = 20 + \log_{10} x + \log_{10} z$
$2\log_{10} y = \log_{10} x + \log_{10} z$
$\log_{10} y = \dfrac{\log_{10} x + \log_{10} z}{2}$
This shows that the AM of $\log_{10} x$ and $\log_{10} z$ is $\log_{10} y$. Thus, Statement II is correct.
From $2\log_{10} y = \log_{10} x + \log_{10} z$, we get:
$\log_{10} y^2 = \log_{10} (xz)$
$y^2 = xz$
The GM of $x$ and $z$ is $\sqrt{xz} = \sqrt{y^2} = y$.
Thus, Statement I is incorrect.
Only Statement II is correct.
Answer: II only
$2(10 + \log_{10} y) = (10 + \log_{10} x) + (10 + \log_{10} z)$
$20 + 2\log_{10} y = 20 + \log_{10} x + \log_{10} z$
$2\log_{10} y = \log_{10} x + \log_{10} z$
$\log_{10} y = \dfrac{\log_{10} x + \log_{10} z}{2}$
This shows that the AM of $\log_{10} x$ and $\log_{10} z$ is $\log_{10} y$. Thus, Statement II is correct.
From $2\log_{10} y = \log_{10} x + \log_{10} z$, we get:
$\log_{10} y^2 = \log_{10} (xz)$
$y^2 = xz$
The GM of $x$ and $z$ is $\sqrt{xz} = \sqrt{y^2} = y$.
Thus, Statement I is incorrect.
Only Statement II is correct.
Answer: II only
Hence, the correct answer is option: 'B'.