<p>If \(S_r = r + \sqrt{r} + \frac{1}{\sqrt{r}} + ... \infty\), \(r > 0\), then which of the following is/are correct.</p>
<p>(a) \(S_6, S_6, S_{12}, S_{20}\) are in AP</p>
<p>(b) \(S_4, S_9, S_{16}\) are irrational</p>
<p>(c) \((2S_4 - 1)^2, (2S_5 - 1)^2, (2S_6 - 1)^2\) are in AP</p>
<p>(d) \(S_2, S_{12}, S_{56}\) are in GP</p>
Step-by-Step Solution
Key Concept: Solve the infinite series equation to get a closed form for $S_r$, then verify each statement by substitution.
<p><strong>Solution:</strong> Given $S_r = r + \sqrt{r} + \frac{1}{\sqrt{r}} + ... \infty$</p><p>Let $S_r = r + Sr$ where the series after r converges.</p><p>$Sr^2 - Sr - r = 0$</p><p>$S_r = \frac{1 + \sqrt{1+4r}}{2}$ (taking the positive root since $r > 0$)</p><p><strong>Alternate (a):</strong> $S_2, S_6, S_{12}, S_{20}$ gives $2, 3, 4, 5$ which are in AP. ✓</p><p><strong>Alternate (b):</strong> $S_4 = \frac{1+\sqrt{17}}{2}, S_9 = \frac{1+\sqrt{37}}{2}, S_{16} = \frac{1+\sqrt{65}}{2}$ are all irrational. ✓</p><p><strong>Alternate (c):</strong> $(2S_4-1)^2, (2S_5-1)^2, (2S_6-1)^2$ gives $17, 21, 25$ which are in AP. ✓</p><p><strong>Alternate (d):</strong> $S_2, S_{12}, S_{56}$ are in GP. ✓</p><p>∴ All options (a), (b), (c), (d) are correct.</p>
Correct Answer: a, b, c, d