Q59 / 155 · WBJEE 2018 · Mathematics

Given that $n$ numbers of arithmetic means are inserted between two sets of numbers $a, 2 b$ and $2 a, b$ where $a, b \in R$. Suppose further that the mih means between these sets of numbers are same, then the ratio $a$ : $b$ equals

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

Solution

$\quad 2 b=(n+2)$ th form $=a+(n+2-1) d$ $2b=a+(n+1) d$ $\Rightarrow \quad d=\frac{b-2 a}{n+1}$ $\therefore m$ th mean $=2 t+m\left(\frac{b-2 a}{n+1}\right)$ and also, $\quad b=(n+2)$ th form $\begin{aligned} &=2 a+(n+2-1) d \\ &=2 a+(n+1) d \\ \Rightarrow \quad d=& \frac{b-2 a}{n+1} \\ \therefore m \text { th mean }=2 a+m\left(\frac{b-2 a}{n+1}\right) \end{aligned}$ According to the question, $a+m\left(\frac{2 b-a}{n+1}\right)=2 a+m\left(\frac{b-2 a}{n+1}\right)$ $\Rightarrow$ $\frac{a}{b}=\frac{m}{n+1-m}$

Hence, the correct answer is option: 'D'.

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