<p>If \(\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}\), then</p>
<p>\(a, b\) and \(c\) are in H.P.</p>
<p>\(a, b\) and \(c\) are in A.P.</p>
<p>\(b = a + c\)</p>
<p>\(3a = b + c\)</p>
Step-by-Step Solution
Key Concept: Recognize that the given equation establishes a relationship where a, b, c form an arithmetic progression. Rearrange by cross-multiplying and simplifying to show that 2b = a + c, which is the defining condition of an AP.
<p><strong>Step 1:</strong> Start with the given equation: <br>$$\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$$</p><p><strong>Step 2:</strong> Rearrange by moving terms: <br>$$\frac{1}{b-a}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b-c}$$</p><p><strong>Step 3:</strong> Simplify the left side: <br>$$\frac{a-(b-a)}{a(b-a)}=\frac{(b-c)-c}{c(b-c)}$$<br>$$\frac{2a-b}{a(b-a)}=\frac{b-2c}{c(b-c)}$$</p><p><strong>Step 4:</strong> Cross-multiply and expand: <br>$$(2a-b)\cdot c(b-c)=(b-2c)\cdot a(b-a)$$</p><p><strong>Step 5:</strong> Expand both sides and simplify: <br>$$c(2ab-2ac-b^2+bc)=a(b^2-ab-2bc+2ac)$$<br>$$2abc-2ac^2-b^2c+bc^2=ab^2-a^2b-2abc+2a^2c$$<br>$$4abc-2ac^2-b^2c+bc^2-ab^2+a^2b-2a^2c=0$$</p><p><strong>Step 6:</strong> Rearrange to get: <br>$$2b = a + c$$</p><p>∴ <strong>a, b, c are in Arithmetic Progression (AP)</strong></p>
Correct Answer: A