Sequence And Series Question 259
Question: If $ m $ is a root of the given equation $ (1-ab)x^{2}- $ $ (a^{2}+b^{2})x $ - $ (1+ab)=0 $ and $ m $ harmonic means are inserted between $ a $ and $ b $ , then the difference between the last and the first of the means equals
Options:
A) $ b-a $
B) $ ab(b-a) $
C) $ a(b-a) $
D) $ ab(a-b) $
Show Answer
Answer:
Correct Answer: B
Solution:
By the given condition $ (1-ab)m^{2}-(a^{2}+b^{2})m-(1+ab)=0 $
$ \Rightarrow $ $ m(a^{2}+b^{2})+(m^{2}+1)ab=m^{2}-1 $ ……(i) Now $ H_1= $ First H.M. between $ a $ and $ b $ $ =\frac{(m+1)ab}{a+mb} $ and $ H_{m}=\frac{(m+1)ab}{b+ma} $
$ \therefore $ $ H_{m}-H_1=(m+1)ab[ \frac{1}{b+ma}-\frac{1}{a+mb} ] $ $ =(m+1)ab\frac{[(m-1)(b-a)]}{(b+ma)(a+mb)} $ $ =\frac{(m^{2}-1)ab(b-a)}{m(a^{2}+b^{2})+(m^{2}+1)ab} $ $ =\frac{(m^{2}-1)ab(b-a)}{m^{2}-1} $ [by (i)] $ =ab(b-a) $ .
BETA
