Question

If $A$ is the AM between $a$ and $b$, then $\frac{(A-2b)}{(A-a)}+\frac{(A-2a)}{(A-b)}=$

• A. $-8$
• B. $2$
• C. $4$
• D. $-4$

Answer 1

The correct option is D

$-4$

Explanation for the correct option:

Step 1. Find the value of $\frac{(A-2b)}{(A-a)}+\frac{(A-2a)}{(A-b)}$

$\because$$A$ is the AM of $a$ and $b$,

$\Rightarrow$$\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}{\mathbf{2}}$

Step 2. Find the value of $\frac{(A-2b)}{(A-a)}$:

$\begin{array}{rcl}\frac{(A-2b)}{(A-a)}& =& \frac{{\displaystyle \frac{\left(a+b\right)}{2}}-2b}{\frac{\left(a+b\right)}{2}-a}\\ & =& \frac{\left(a+b\right)-4b}{\left(a+b\right)-2a}\end{array}$

$\begin{array}{rcl}\frac{\left(A-2b\right)}{\left(A-a\right)}& =& \frac{\left(a-2b\right)}{\left(b-a\right)}\end{array}$ …..(1)

Step 3. Find the value of $\frac{(A-2a)}{(A-b)}$:

$\begin{array}{rcl}\frac{(A-2a)}{(A-b)}& =& \frac{{\displaystyle \frac{\left(a+b\right)}{2}}-2a}{\frac{\left(a+b\right)}{2}-b}\\ & =& \frac{\left(a+b\right)-4a}{\left(a+b\right)-2b}\end{array}$

$\begin{array}{rcl}\frac{\left(A-2a\right)}{\left(A-b\right)}& =& \frac{\left(b-3a\right)}{a-b}\end{array}$ …..(2)

Step 4. Add equation (i) and (ii):

$\begin{array}{rcl}\frac{(A-2b)}{(A-a)}+\frac{(A-2a)}{(A-b)}& =& \frac{(a-3b)}{(b-a)}+\frac{(b-3a)}{(a-b)}\\ & =& \frac{(a-3b-b+3a)}{(b-a)}\\ & =& \frac{(4a-4b)}{(b-a)}\\ & =& \frac{-4(b-a)}{(b-a)}\\ & =& \mathbf{-}\mathbf{4}\end{array}$

Hence, Option ‘D’ is Correct.