Question

Given that $n$ $A{M}^{\prime }s$ are inserted between two sets of numbers $a,2b$ and $2a,b$ where $a,b\in R.$ Suppose further that $mth$ mean between these sets of numbers is same, then the ratio $a:b$ is equal to

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

Answer 1

The correct option is D $m:(n-m+1)$

Let the common difference be d. As there are $nA{M}^{\prime }s$ between $a$ and $b$ and total number of terms in the sequence is $=n+2$

$\Rightarrow nthtermb=a+(n-1)d$

$d=\frac{(b-a)}{n+1}$

so, 2nd term that is first term $A\left(1\right)=a+\left[\frac{(b-a)}{(n+1)}\right]$

3rd term that is second mean $A\left(2\right)=a+[2\times \frac{(b-a)}{(n+1)}]$

In the way $r^{th}$ mean $=a+[r\times \frac{(b-a)}{(n+1)}]$

In the first sequence first term is a $n^{th}$ term $=2b$ and $nA{M}^{\prime }s$ between them.

As such from above concept -

$\Rightarrow r^{th}$ term is $a+[r\times \frac{(2b-a)}{(n+1)}]$

$m^{th}$ term is $a+[m\times \frac{(2b-a)}{(n+1)}]$

$ii)$ Similarly for second sequence -

$m^{th}$ mean $=2a+[m\times \frac{(b-2a)}{(n+1)}]$

$iii).$ Since the $m^{th}$ mean, are equal equation like the above.

$=a+[m\times \frac{(2b-a)}{(n+1)}]$

$=2a+m\times \frac{(b-a)}{(n+1)}$

$=[m\times \frac{(2b-a)}{(n+1)}]-[m\times \frac{(b-2a)}{(n+1)}]$

$=2a-a$

$\Rightarrow \frac{m}{(n+1)}\times (2b-a-b+2a)=a$

$\frac{a+b}{a}=\frac{n+1}{m}$

subtracting on both sides

$\Rightarrow \frac{a+b}{a}-1=\frac{n+1}{m}-1$

$\Rightarrow \frac{a+b-a}{a}=\frac{n+1-m}{m}$

$\Rightarrow \frac{b}{a}=\frac{n+1-m}{m}$

$\Rightarrow \frac{a}{b}=\frac{m}{n-m+1}$

Hence, the answer is $\frac{m}{n-m+1}.$