Question:

Step 1: Identify the common difference of the arithmetic series. Since the first three terms of the arithmetic series are $a$, $b$, and $a^2$, the common difference must be constant. Therefore: $$b - a = a^2 - b$$ Solving for $b$: $$2b = a + a^2$$ $$b = \frac{a + a^2}{2}$$ Step 2: Determine the common ratio of the geometric series. The geometric series has first three terms $a$, $a^2$, and $b$. The common ratio is: $$r = \frac{a^2}{a} = a$$ For this to be a valid geometric series, the ratio between consecutive terms must be constant: $$\frac{b}{a^2} = r = a$$ Step 3: Set up an equation using the geometric series condition. Substitute the expression for $b$ from Step 1 into the condition from Step 2: $$\frac{b}{a^2} = a$$ $$\frac{\frac{a + a^2}{2}}{a^2} = a$$ $$\frac{a + a^2}{2a^2} = a$$ $$\frac{1 + a}{2a} = a$$ Multiplying both sides by $2a$: $$1 + a = 2a^2$$ $$2a^2 - a - 1 = 0$$ Step 4: Solve the quadratic equation. Factoring: $$(2a + 1)(a - 1) = 0$$ This gives us: $$a = -\frac{1}{2} \quad \text{or} \quad a = 1$$ Since we are told that $a$ is negative: $$a = -\frac{1}{2}$$ Step 5: Calculate the value of $b$. Using $b = \frac{a + a^2}{2}$ with $a = -\frac{1}{2}$: $$b = \frac{-\frac{1}{2} + \frac{1}{4}}{2} = \frac{-\frac{1}{4}}{2} = -\frac{1}{8}$$ Step 6: Find the sum of the infinite geometric series. The geometric series has: - First term: $a = -\frac{1}{2}$ - Common ratio: $r = a = -\frac{1}{2}$ Since $|r| = \frac{1}{2} < 1$, the series converges and its sum is: $$S = \frac{a}{1 - r} = \frac{-\frac{1}{2}}{1 - (-\frac{1}{2})} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{2} \times \frac{2}{3} = -\frac{1}{3}$$ **Final Answer:** The sum of the infinite geometric series is $\boxed{-\frac{1}{3}}$, which corresponds to **Option 3**.