Sequences & Series
Arithmetic Progression
Grade None
Question:
<p>If \(a^2 + 2bc, b^2 + 2ca, c^2 + 2ab\) are in A.P., then</p>
<p>\((a-b)(c-a), (a-b)(b-c), (b-c)(c-a)\) are in A.P.</p>
<p>\(b-c, c-a, a-b\) are in H.P.</p>
<p>\(a+b, b+c, c+a\) are in A.P.</p>
<p>\(a^2, b^2, c^2\) are in H.P.</p>
Step-by-Step Solution
Key Concept: If three expressions are in A.P., then twice the middle term equals the sum of the first and third terms. Apply this condition and simplify using algebraic identities to find the relationship between a, b, c.
<p><strong>Step 1:</strong> Apply the A.P. condition: twice the middle term equals sum of first and third.</p><p>2(b² + 2ca) = (a² + 2bc) + (c² + 2ab)</p><p><strong>Step 2:</strong> Expand and simplify.</p><p>2b² + 4ca = a² + 2bc + c² + 2ab</p><p>2b² + 4ca - a² - 2bc - c² - 2ab = 0</p><p><strong>Step 3:</strong> Rearrange terms strategically.</p><p>-(a² - 2ab + b²) - (b² - 2bc + c²) - (c² - 2ca + a²) = 0</p><p>-(a - b)² - (b - c)² - (c - a)² = 0</p><p><strong>Step 4:</strong> Since sum of squares equals zero, each square must be zero.</p><p>(a - b)² = 0, (b - c)² = 0, (c - a)² = 0</p><p>∴ <strong>a = b = c</strong> (or a, b, c are equal / the terms are all identical)</p>
Correct Answer: AB