<p><strong>33.</strong> If \(a, b\) and \(c\) are in G.P. and \(x, y\), respectively, are the arithmetic means between \(a, b\) and \(b, c\), then the value of \(\dfrac{a}{x} + \dfrac{c}{y}\) is</p>
Step-by-Step Solution
Key Concept: When x is the arithmetic mean between a and b, then x = (a+b)/2. Similarly, y = (b+c)/2. Since a, b, c are in G.P., we have b² = ac. Use these relationships to simplify the given expression.
<p><strong>Step 1:</strong> Since x is the arithmetic mean between a and b:</p><p>x = (a + b)/2, so a + b = 2x</p><p><strong>Step 2:</strong> Since y is the arithmetic mean between b and c:</p><p>y = (b + c)/2, so b + c = 2y</p><p><strong>Step 3:</strong> Since a, b, c are in G.P., we have:</p><p>b² = ac</p><p><strong>Step 4:</strong> Calculate a/x + c/y:</p><p>a/x + c/y = a/[(a+b)/2] + c/[(b+c)/2]</p><p>= 2a/(a+b) + 2c/(b+c)</p><p><strong>Step 5:</strong> From a + b = 2x, we get a = 2x - b. From b + c = 2y, we get c = 2y - b.</p><p>Alternatively, using b² = ac:</p><p>a/x + c/y = 2a/(a+b) + 2c/(b+c)</p><p>Multiply first fraction by b/b: 2ab/[b(a+b)] and second by b/b: 2bc/[b(b+c)]</p><p><strong>Step 6:</strong> Since b² = ac, we have ab + bc = b(a+c) and the common denominator approach:</p><p>= 2a(b+c) + 2c(a+b) / [(a+b)(b+c)]</p><p>= 2[ab + ac + ac + bc] / [(a+b)(b+c)]</p><p>= 2[ab + 2ac + bc] / [(a+b)(b+c)]</p><p>Since b² = ac: = 2[ab + 2b² + bc] / [(a+b)(b+c)] = 2b(a + 2b + c) / [(a+b)(b+c)]</p><p><strong>Step 7:</strong> Simplify: = 2b[(a+b) + (b+c)] / [(a+b)(b+c)] = 2b(a+b)/(a+b)(b+c) + 2b(b+c)/(a+b)(b+c)</p><p>= 2b/(b+c) + 2b/(a+b) = 2b[1/(b+c) + 1/(a+b)] = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: B