Q256 / 270 · NDA 2024 (Phase 1) · Mathematics

If $a, b, c$ are in HP, then what is $\frac{1}{b-a}+\frac{1}{b-c}$ equal to? 1. $\frac{2}{b}$ 2. $\frac{1}{a}+\frac{1}{c}$ 3. $\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ Select the correct answer using the code given below:

  • A
    1 only
  • Correct answer
    B
    2 only
  • C
    3 only
  • D
    1,2 and 3

Solution

$\because a, b, c$ are in H.P. $\Rightarrow \quad \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ $\begin{aligned} & \Rightarrow \quad b=\frac{2 a c}{a+c} \\ & b-a=\frac{2 a c}{a+c}-a=\frac{2 a c-a^2-a c}{a+c} \\ & =\frac{a c-a^2}{a+c}=\frac{a(c-a)}{a+c} \\ & \Rightarrow \quad \frac{1}{b-a}=\frac{a+c}{-a(a-c)} \\ & \text { Now, } \quad b-c=\frac{2 a c}{a+c}-c=\frac{2 a c-a c-c^2}{a+c} \\ & =\frac{a c-c^2}{a+c}=\frac{c(a-c)}{a+c} \\ & \Rightarrow \quad \frac{1}{b-c}=\frac{a+c}{c(a-c)} \\ & \therefore \quad \frac{1}{b-a}+\frac{1}{b-c}=\frac{a+c}{-a(a-c)}+\frac{a+c}{c(a-c)} \\ & =\frac{(a+c)}{(a-c)}\left[-\frac{1}{a}+\frac{1}{c}\right] \\ & =\frac{a+c}{a c}=\frac{1}{a}+\frac{1}{c} \\ & =\frac{2}{b}=\frac{1}{b}+\frac{1}{b}=\frac{1}{b}+\frac{1}{2}\left(\frac{1}{a}+\frac{1}{c}\right) \end{aligned}$ Hence, only statement (1) and (2) are correct. Statement (3) is not correct.

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

Relevant Questions