1. **State the problem:** We are given the first two terms of a sequence: $a_1 = 9$ and $a_2 = -3$. We need to find $a_3 + b_3$, where $a_3$ is the third term of the arithmetic sequence and $b_3$ is the third term of the geometric sequence formed by these terms. 2. **Recall formulas:** - For an arithmetic sequence, the $n$th term is given by: $$a_n = a_1 + (n-1)d$$ where $d$ is the common difference. - For a geometric sequence, the $n$th term is: $$b_n = b_1 \cdot r^{n-1}$$ where $r$ is the common ratio. 3. **Find the arithmetic sequence third term $a_3$:** - Calculate the common difference: $$d = a_2 - a_1 = -3 - 9 = -12$$ - Then: $$a_3 = a_1 + 2d = 9 + 2(-12) = 9 - 24 = -15$$ 4. **Find the geometric sequence third term $b_3$:** - Calculate the common ratio: $$r = \frac{a_2}{a_1} = \frac{-3}{9} = -\frac{1}{3}$$ - Then: $$b_3 = b_1 \cdot r^2 = 9 \cdot \left(-\frac{1}{3}\right)^2 = 9 \cdot \frac{1}{9} = 1$$ 5. **Calculate $a_3 + b_3$:** $$a_3 + b_3 = -15 + 1 = -14$$ **Final answer:** $$\boxed{-14}$$