Sequence And Series Question 3
Question: If $ a,\ b,\ c $ are three distinct positive real numbers which are in H.P., then $ \frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b} $ is
Options:
A) Greater than or equal to 10
B) Less than or equal to 10
C) Only equal to 10
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ \frac{1}{a},\ \frac{1}{b},\ \frac{1}{c} $ are in A.P. Let $ \frac{1}{a}=p-q,\ \frac{1}{b}=p $ and $ \frac{1}{c}=p+q $ , where $ p,\ q>0 $ and $ p>\ q $ . Now, substitute these values in $ \frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b} $ then it reduces to $ 10+\frac{14q^{2}}{p^{2}-q^{2}} $ which is obviously greater than $ 10(as\ p>q>0) $ . Trick: Put $ a=1,\ b=\frac{1}{2},\ c=\frac{1}{3} $ . The expression has the value $ \frac{3+1}{2-\frac{1}{2}}+\frac{1+1}{\frac{2}{3}-\frac{1}{2}}=\frac{8}{3}+12>10 $ .
BETA
