<p>Three numbers form an increasing GP. If the middle number is doubled, then the new numbers are in AP. The common ratio of the GP is</p>
<p>(a) \(2 - \sqrt{3}\)</p>
<p>(b) \(2 + \sqrt{3}\)</p>
<p>(c) \(\sqrt{3} - 2\)</p>
<p>(d) \(3 + 2\)</p>
Step-by-Step Solution
Key Concept: Express the three GP terms, use the AP condition for the doubled middle term, and solve the resulting quadratic equation.
<p><strong>Solution:</strong></p><p>Let the three numbers in GP be $\frac{a}{r}, a, ar$.</p><p>Since the numbers form an increasing GP, $r > 1$.</p><p>When the middle number is doubled, the numbers $\frac{a}{r}, 2a, ar$ are in AP.</p><p>For AP: $4a = \frac{a}{r} + ar$</p><p>Dividing by $a$: $4 = \frac{1}{r} + r$</p><p>$r^2 - 4r + 1 = 0$</p><p>$r = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$</p><p>Since $r > 1$ and $2 - \sqrt{3} < 1$, we have $r = 2 + \sqrt{3}$</p><p>∴ Answer is (b)</p>
Correct Answer: b