<p>Suppose <i>x</i> and <i>y</i> are two real numbers such that the <i>r</i>th mean between <i>x</i> and <i>2y</i> is equal to <i>r</i>th mean between <i>2x</i> and <i>y</i>, when <i>n</i> arithmetic means are inserted between them in both the cases. Then \(\frac{n+1}{r} - \frac{y}{x}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the formula for the rth arithmetic mean inserted between two numbers and equate the two conditions to find the relationship between n, r, x, and y.
<p><strong>Solution:</strong> Let $A_1, A_2, A_3, \ldots, A_n$ be <i>n</i> arithmetic means between two numbers <i>a</i> and <i>b</i>. Then $a, A_1, A_2, A_3, \ldots, A_r, \ldots, A_n, b$ are in AP.</p><p>Let <i>d</i> be the common difference, then:</p><p>$$d = \frac{b - a}{n + 1}$$</p><p>For the first case (between <i>x</i> and <i>2y</i>):</p><p>$$A_r^{(1)} = x + r \cdot \frac{2y - x}{n + 1}$$</p><p>For the second case (between <i>2x</i> and <i>y</i>):</p><p>$$A_r^{(2)} = 2x + r \cdot \frac{y - 2x}{n + 1}$$</p><p>Given that $A_r^{(1)} = A_r^{(2)}$:</p><p>$$x + r \cdot \frac{2y - x}{n + 1} = 2x + r \cdot \frac{y - 2x}{n + 1}$$</p><p>$$x - 2x = r \cdot \frac{y - 2x - 2y + x}{n + 1}$$</p><p>$$-x = r \cdot \frac{-y - x}{n + 1}$$</p><p>$$x(n + 1) = r(x + y)$$</p><p>$$\frac{n+1}{r} = \frac{x + y}{x} = 1 + \frac{y}{x}$$</p><p>$$\therefore \frac{n+1}{r} - \frac{y}{x} = 1$$</p><p>∴ Answer is (b) 1.</p>
Correct Answer: B