<p>The corresponding first and the \((2n-1)\)th terms of an A.P., a G.P. and a H.P. are equal. If their \(n\)th terms are \(a, b\) and \(c\), respectively, then</p>
Step-by-Step Solution
Key Concept: When first and (2n-1)th terms are equal in A.P., G.P., and H.P., use the property that the nth term is the geometric mean of equidistant terms. For A.P.: a = (first + (2n-1)th)/2; for G.P.: b² = first × (2n-1)th; apply the same logic to relate a, b, c.
<p><strong>Step 1:</strong> Let the common first term and (2n-1)th term be equal to some value α for all three progressions.</p><p><strong>Step 2:</strong> For A.P. with nth term = a: The property states that if first and (2n-1)th terms are equal, then a = [first + (2n-1)th]/2 = α</p><p><strong>Step 3:</strong> For G.P. with nth term = b: Using b² = first × (2n-1)th = α × α, we get b² = α²</p><p><strong>Step 4:</strong> For H.P. with nth term = c: If the H.P. has nth term c, then 1/c is the nth term of corresponding A.P. For an H.P., when equidistant terms are equal: 1/c = [1/first + 1/(2n-1)th]/2 = [1/α + 1/α]/2 = 1/α</p><p><strong>Step 5:</strong> Therefore: a = α, b² = α², and c = α, which gives us b² = α² and a, c are the outer terms.</p><p><strong>Step 6:</strong> From the relationship between A.P., G.P., H.P.: <strong>a + c = 2b</strong> (since a and c are related through the harmonic-arithmetic mean relationship)</p><p>∴ Answer: <strong>B (a + c = 2b)</strong></p>
Correct Answer: B