Question:

Step 1: Identify the convergence condition for an infinite geometric series. For an infinite geometric series with first term $a_1$ and common ratio $r$ to converge, we require: $$|r| < 1$$ Step 2: Write the formula for the sum of a convergent infinite geometric series. The sum of a convergent infinite geometric series is given by: $$S = \frac{a_1}{1-r}$$ Step 3: Apply the condition that the sum is negative. We are given that the sum is negative, so: $$\frac{a_1}{1-r} < 0$$ This inequality tells us that the numerator $a_1$ and denominator $(1-r)$ must have opposite signs. Step 4: Analyze Case 1 where $a_1 > 0$. If $a_1 > 0$, then for the fraction to be negative, we need: $$1 - r < 0 \implies r > 1$$ However, this contradicts the convergence condition $|r| < 1$. Therefore, this case is impossible. Step 5: Analyze Case 2 where $a_1 < 0$. If $a_1 < 0$, then for the fraction to be negative, we need: $$1 - r > 0 \implies r < 1$$ Combined with the convergence condition $|r| < 1$ (which means $-1 < r < 1$), this case is compatible. We have $-1 < r < 1$ and $r < 1$, which is consistent. Step 6: Conclude what must always be true. From the analysis above, we must have: $$a_1 < 0 \text{ and } -1 < r < 1$$ Step 7: Evaluate each option. **Option 1:** $a_1 < 0$ — This is always true based on our analysis. ✓ **Option 2:** $a_1 > 0$ or $r < 0$ — This is not always true. For example, if $a_1 = -1$ and $r = 0.5$, then $S = \frac{-1}{1-0.5} = -2 < 0$. Here, $a_1 > 0$ is false and $r < 0$ is false, making the entire statement false. ✗ **Option 3:** $a_1 < 0$ and $r < 0$ — This is not always true. The counterexample $a_1 = -1$ and $r = 0.5$ shows that $r$ can be positive. ✗ **Option 4:** $r < 0$ — This is not always true, as shown by the counterexample above. ✗ **Final Answer:** Option 1 is always true. The condition $a_1 < 0$ must hold for any infinite geometric sequence with a convergent and negative sum.