Question

If n arithmetic means are inserted between two sets of numbers $ a,2b $ and $ 2a,b $ where $ a,b\in R $ suppose mth mean between these two sets of numbers is same, then the ratio $ a:b $ equals to

• A. $ n-m+1:m $
• B. $ n-m+1:n $
• C. $ m:n-m+1 $
• D. $ n:n-m+1 $

Answer 1

Answer: (C)

Since, n arithmetic means are inserted between a and 2b, $ \therefore $ $ {{a}_{m}} $ (mth mean) $ =a+\frac{m}{n+1}(2b-a) $ and n arithmetic means are inserted between 2a and b. $ \therefore $ $ a{{}_{m}} $ (mth mean) $ =2a+\frac{m}{n+1}(b-2a) $ Given, $ {{a}_{m}}=a{{}_{m}} $ $ \Rightarrow $ $ a+\frac{m}{n+1}.(2b-a)=2a+\frac{m}{(n+1)}(b-2a) $ $ \Rightarrow $ $ \frac{a}{b}=\frac{m}{n-m+1} $ $ \therefore $ $ a:b=m:n-m+1 $