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Sequence and Series

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Question : 33 of 86

Marks: +1, -0

If

\frac{1}{x+y}, \frac{1}{2y}, \frac{1}{y+z}

are in A.P., then

x, y, z

are in

A.P.
G.P.
Cannot determined
None

Solution:

Concept:

If three terms are in Arithmetic Progression (A.P.), then twice the middle term equals the sum of the first and third terms.

Explanation:

Given

\frac{1}{x+y},\ \frac{1}{2y},\ \frac{1}{y+z}

are in A.P.

So,

2 \times \frac{1}{2y} = \frac{1}{x+y} + \frac{1}{y+z}

.

This simplifies to

\frac{1}{y} = \frac{1}{x+y} + \frac{1}{y+z}

.

Combine RHS:

\frac{1}{y} = \frac{(y+z)+(x+y)}{(x+y)(y+z)} = \frac{x+2y+z}{(x+y)(y+z)}

.

Cross-multiply:

(x+y)(y+z) = y(x+2y+z)

.

Expand:

xy + xz + y^2 + yz = xy + 2y^2 + yz

.

Cancel

xy

and

yz

:

xz + y^2 = 2y^2 \Rightarrow xz = y^2

.

Thus

y^2 = xz

, which means

x, y, z

are in Geometric Progression (G.P.).

Answer:

x, y, z

are in G.P.

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