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, $a_m=a{}_m$ $\Rightarrow$ $a+\frac{m}{n+1}.(2b-a)=2a+\frac{m}{(n+1)}(b-2a)$ $\Rightarrow$ $\frac{a}{b}=\frac{m}{n-m+1}$ $\therefore$ $a:b=m:n-m+1$