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Sequence and Series

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Question : 86 of 86

Marks: +1, -0

If

a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{2}}

a, b, c

x, y, z

are in

AP
GP
HP
AGP

Solution:

Concept:

If

a, b, c

are in GP then

b^2 = ac

.

Explanation:

Let

a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k

.

Then

a = k^x

,

b = k^y

,

c = k^z

.

Since

a, b, c

are in GP, we have

b^2 = ac

.

Substitute:

(k^y)^2 = k^x \cdot k^z

⟹

k^{2y} = k^{x+z}

.

Equating exponents:

2y = x+z

.

Thus

x, y, z

are in arithmetic progression (AP).

Answer:

A. AP

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