Test Index

Sequence and Series

© examsnet.com

Question : 65 of 86

Marks: +1, -0

If A be the A . M of two positive unequal quantities

x

y

G

G \cdot M,

then

$A < G$
$A > G$
$A \geq G$
$A \leq G$

Solution:

Concept:

For two positive unequal numbers, the arithmetic mean (A) is always greater than the geometric mean (G).

Explanation:

Given

A = \frac{x+y}{2}

and

G = \sqrt{xy}

.

Since

x \neq y

and both positive, consider the identity

(x-y)^2 > 0

.

Expanding:

x^2 - 2xy + y^2 > 0

implies

x^2 + y^2 > 2xy

.

Add

2xy

to both sides:

x^2 + 2xy + y^2 > 4xy

i.e.

(x+y)^2 > 4xy

.

Taking square roots (positive):

x+y > 2\sqrt{xy}

i.e.

\frac{x+y}{2} > \sqrt{xy}

.

Thus

A > G

.

Answer:

A > G

© examsnet.com

Go to Question:

Prev Question

Next Question