Question:

Step 1: Set up the geometric sequence with variables. Let the first term be $a$ and the common ratio be $r$, where $|r| < 1$ (required for convergence of an infinite geometric series). Step 2: Express the sum formula and use the given condition. The sum of an infinite geometric series is: $$S = \frac{a}{1-r}$$ We are given that the second term equals 1: $$ar = 1$$ From this, we can express the first term in terms of $r$: $$a = \frac{1}{r}$$ Step 3: Substitute to express $S$ in terms of $r$ only. Substituting $a = \frac{1}{r}$ into the sum formula: $$S = \frac{1/r}{1-r} = \frac{1}{r(1-r)}$$ Step 4: Determine the valid range for $r$. Since we need $S > 0$, the denominator must be positive: $$r(1-r) > 0$$ This inequality is satisfied when $0 < r < 1$. Step 5: Convert the minimization problem. To minimize $S = \frac{1}{r(1-r)}$, we need to maximize the denominator $f(r) = r(1-r)$ for $0 < r < 1$. Expanding: $$f(r) = r - r^2$$ Step 6: Find the critical point using calculus. Taking the derivative with respect to $r$: $$f'(r) = 1 - 2r$$ Setting $f'(r) = 0$: $$1 - 2r = 0 \implies r = \frac{1}{2}$$ Step 7: Verify this is a maximum and calculate the maximum value. The second derivative is $f''(r) = -2 < 0$, confirming this is a maximum. The maximum value of $f(r)$ is: $$f\left(\frac{1}{2}\right) = \frac{1}{2} \cdot \left(1 - \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$ Step 8: Calculate the minimum value of $S$. The minimum value of $S$ is: $$S_{\min} = \frac{1}{f_{\max}} = \frac{1}{1/4} = 4$$ **Final Answer:** The minimum possible value of $S$ is $\boxed{4}$, which corresponds to **Option 2**.