Sequence And Series Question 524
Question: A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $ \frac{13}{12} $ , then the numbers are
Options:
A) $ \frac{1}{4},\ \frac{4}{1} $
B) $ \frac{3}{4},\ \frac{4}{3} $
C) $ \frac{2}{5},\ \frac{5}{2} $
D) $ \frac{3}{2},\ \frac{2}{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
Suppose that required numbers $ a $ and $ b $ . Therefore according to the conditions $ a=\frac{1}{b} $ and $ \frac{a+b}{2}=\frac{13}{12} $
$ \Rightarrow $ $ a+b=\frac{13}{6} $
$ \Rightarrow $ $ a+\frac{1}{a}=\frac{13}{6}\Rightarrow 6a^{2}-13a+6=0 $
$ \Rightarrow $ $ ( a-\frac{3}{2} ),( a-\frac{2}{3} )=0 $
$ \Rightarrow $ $ a=\frac{3}{2} $ and $ b=\frac{2}{3} $ or $ a=\frac{2}{3} $ and $ b=\frac{3}{2} $ . Trick: Find the A.M. of option (a), (b), (c), (d) one by one.
BETA
