2. Let $\left\{a_{n}\right\}_{n \geq 1}$ be an arithmetic sequence and $\left\{g_{n}\right\}_{n \geq 1}$ be a geometric sequence such that the first four terms of $\left\{\bar{a}_{n}+g_{n}\right\}$ are $0,0,1$, and 0 , in that order. What is the 10 th term of $\left\{a_{n}+g_{n}\right\}$ ?
Answer: $\quad-54$ Let the terms of the geometric sequence be $a, r a, r^{2} a, r^{3} a$. Then, the terms of the arithmetic sequence are $-a,-r a,-r^{2} a+1,-r^{3} a$. However, if the first two terms of this sequence are $-a,-r a$, the next two terms must also be $(-2 r+1) a,(-3 r+2) a$. It is clear that $a \neq 0$ because $a_{3}+g_{3} \neq 0$, so $-r^{3}=-3 r+2 \Rightarrow r=1$ or -2 . However, we see from the arithmetic sequence that $r=1$ is impossible, so $r=-2$. Finally, by considering $a_{3}$, we see that $-4 a+1=5 a$, so $a=1 / 9$. We also see that $a_{n}=(3 n-4) a$ and $g_{n}=(-2)^{n-1} a$, so our answer is $a_{10}+g_{10}=(26-512) a=-486 a=-54$.