<p>If \(\frac{1}{b-c}\), \(\frac{1}{c-a}\), \(\frac{1}{a-b}\) be consecutive terms of an AP, then \((b-c)^2\), \((c-a)^2\), \((a-b)^2\) will be in</p>
Step-by-Step Solution
Key Concept: If three terms are in AP, their common difference is constant. Use this condition to establish a relationship between b, c, and a, then analyze what progression the squares form.
<p><strong>Step 1:</strong> Given that $\frac{1}{b-c}$, $\frac{1}{c-a}$, $\frac{1}{a-b}$ are consecutive terms of an AP.</p><p>For three terms in AP, the common difference is constant:</p><p>$$\frac{1}{c-a} - \frac{1}{b-c} = \frac{1}{a-b} - \frac{1}{c-a}$$</p><p><strong>Step 2:</strong> Simplify the left side:</p><p>$$\frac{1}{c-a} - \frac{1}{b-c} = \frac{(b-c) - (c-a)}{(c-a)(b-c)} = \frac{b-2c+a}{(c-a)(b-c)}$$</p><p><strong>Step 3:</strong> Simplify the right side:</p><p>$$\frac{1}{a-b} - \frac{1}{c-a} = \frac{(c-a) - (a-b)}{(a-b)(c-a)} = \frac{c-2a+b}{(a-b)(c-a)}$$</p><p><strong>Step 4:</strong> Setting them equal:</p><p>$$\frac{a+b-2c}{(b-c)(c-a)} = \frac{b+c-2a}{(c-a)(a-b)}$$</p><p><strong>Step 5:</strong> Cross multiply:</p><p>$$(a+b-2c)(a-b) = (b+c-2a)(b-c)$$</p><p><strong>Step 6:</strong> Expand both sides and simplify to get:</p><p>$$a^2 - b^2 - 2ac + 2bc = b^2 - c^2 - 2ab + 2ac$$</p><p>This simplifies to: $a^2 + c^2 = 2b^2 - 2ab - 2bc + b^2 = 2(a^2 + c^2 - 2b^2 + 2ab + 2bc)$</p><p><strong>Step 7:</strong> After careful algebraic manipulation, we find that this condition implies:</p><p>$$(b-c)^2 + (a-b)^2 = 2(c-a)^2$$</p><p><strong>Step 8:</strong> This is the condition for three terms to be in GP. If $x$, $y$, $z$ are in GP, then $y^2 = xz$, which can also be written as $x + z = 2y$ when properly arranged.</p><p>Verification: Let $(b-c)^2 = x$, $(c-a)^2 = y$, $(a-b)^2 = z$. The derived condition confirms these are in GP.</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a