Sequence And Series Question 325
Question: If A.M and G.M of x and y are in the ratio p : q, then x : y is
[Kerala (Engg.) 2005]
Options:
A) $ p-\sqrt{p^{2}+q^{2}} $ : $ p+\sqrt{p^{2}+q^{2}} $
B) $ p+\sqrt{p^{2}-q^{2}} $ : $ p-\sqrt{p^{2}-q^{2}} $
C) $ p:q $
D) $ p+\sqrt{p^{2}+q^{2}} $ : $ p-\sqrt{p^{2}+q^{2}} $
E) $ q+\sqrt{p^{2}-q^{2}} $ : $ q-\sqrt{p^{2}-q^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\frac{x+y}{2}}{\sqrt{xy}}=\frac{p}{q} $ $ \frac{x+y}{2(\sqrt{xy})}=\frac{p}{q} $ ?..(i) $ \frac{x^{2}+y^{2}+2xy}{4xy}=\frac{p^{2}}{q^{2}} $ $ \frac{x^{2}+y^{2}+2xy-4xy}{4xy}=\frac{p^{2}-q^{2}}{q^{2}} $ $ \frac{{{(x-y)}^{2}}}{4xy}=\frac{p^{2}-q^{2}}{q^{2}} $ $ \frac{x-y}{2\sqrt{xy}}=\frac{\sqrt{p^{2}-q^{2}}}{q} $ ?..(ii) Equation (i) is divided by (ii), Then $ \frac{x+y}{x-y}=\frac{p}{\sqrt{p^{2}-q^{2}}} $ ; $ \frac{x}{y}=\frac{p+\sqrt{p^{2}-q^{2}}}{p-\sqrt{p^{2}-q^{2}}} $ .
BETA
