HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

Let $\{a_{n}\}_{n \geq}$ 1} $be an arithmetic sequence and \{g_{n}\}_{n \geq 1}$ be a geometric sequence such that the first four terms of \{a_{n}+g_{n}\} $are$ 0,0,1 $, and 0 , in that order. What is the 10th term of \{a_{n}+g_{n}\}$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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 $a_{10}+g_{10}=(26-512) a=-486 a=-54$ .

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