(A) Let $a^{1/x} = b^{1/y} = c^{1/z} = k$ (where $k \neq 1$).Then $a = k^x, b = k^y, c = k^z$.Since $a, b, c$ are in geometric progression,we have $b^

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(A) Let $a^{1/x} = b^{1/y} = c^{1/z} = k$ (where $k \neq 1$).Then $a = k^x, b = k^y, c = k^z$.Since $a, b, c$ are in geometric progression,we have $b^2 = ac$.Substituting the values,we get $(k^y)^2 = k^x \cdot k^z$.$k^{2y} = k^{x+z}$.Equating the exponents,$2y = x + z$.This implies that $x, y, z$ are in arithmetic progression.

Class 11 Mathematics Sequences and Series Relation between A.P., G.P.

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If $a^{1/x} = b^{1/y} = c^{1/z}$ and $a, b, c$ are in geometric progression,then $x, y, z$ are in.....

A
Arithmetic progression.
B
Geometric progression.
C
Harmonic progression.
D
None of these.

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Relation between A.P., G.P.

The product of $n$ positive numbers is unity. Their sum is

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If $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}$ are in $A.P.$,then $a, \frac{1}{b}, c$ are in

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If the first and $(2n - 1)^{th}$ terms of an $A.P.$,$G.P.$,and $H.P.$ are equal and their $n^{th}$ terms are respectively $a, b$ and $c$,then:

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If $a^{1/x} = b^{1/y} = c^{1/z}$ and $a, b, c$ are in geometric progression,then $x, y, z$ are in.....

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