Correct Solution: Option A
Step-by-step Solution:
MCQ Solution
Q. If \(a, b, c\) are in Geometric Progression and
\( a^{\tfrac{1}{x}} = b^{\tfrac{1}{y}} = c^{\tfrac{1}{z}} \),
then \(x, y, z\) are in
(a) Arithmetic Progression
(b) Geometric Progression
(c) \( \tfrac{2}{y} = \tfrac{1}{x} + \tfrac{1}{z} \)
(d) \( x = y+z \)
Solution:
Since \(a, b, c\) are in Geometric Progression (G.P.),
we can write:
\[
b^2 = ac
\]
Now given:
\[
a^{\tfrac{1}{x}} = b^{\tfrac{1}{y}} = c^{\tfrac{1}{z}} = k \quad (\text{say})
\]
Taking logarithms:
\[
\frac{1}{x}\ln a = \frac{1}{y}\ln b = \frac{1}{z}\ln c
\]
⇒
\[
\ln a : \ln b : \ln c = \tfrac{1}{x} : \tfrac{1}{y} : \tfrac{1}{z}
\]
But since \(a, b, c\) are in G.P.,
\[
\ln a, \ln b, \ln c
\]
are in **A.P.** (because logarithm of G.P. terms forms an A.P.).
Hence, the reciprocals \( \tfrac{1}{x}, \tfrac{1}{y}, \tfrac{1}{z} \) are also in A.P.
Which implies:
\[
x, y, z \ \text{are in Arithmetic Progression}.
\]
✅ Correct Answer: (a) Arithmetic Progression