Question

A sequence of positive integers with $a_1 = 1$ and $a_9 + a_{10} = 646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n \ge 1,$ the terms $a_{2n-1},$ $a_{2n},$ and $a_{2n+1}$ are in geometric progression, and the terms $a_{2n},$ $a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000.$ Find $n + a_n.$

Answer

Solution:

Let $a_2 = r.$ The geometric condition gives $a_3 = r^2,$ the arithmetic condition gives $a_4 = 2r^2 - r = r(2r-1),$ then $a_5 = (2r-1)^2,$ and so on: inductively $a_{2k+1} = \bigl(kr - (k-1)\bigr)^2, \qquad a_{2k+2} = \bigl(kr - (k-1)\bigr)\bigl((k+1)r - k\bigr).$ In particular $a_9 = (4r-3)^2$ and $a_{10} = (4r-3)(5r-4),$ so $a_9 + a_{10} = (4r-3)(9r-7) = 646.$ Expanding gives $36r^2 - 55r - 625 = 0,$ which factors as $(r - 5)(36r + 125) = 0,$ so $r = 5.$

With $r = 5$ we get $kr - (k-1) = 4k + 1,$ so $a_{2k+1} = (4k+1)^2$ and $a_{2k+2} = (4k+1)(4k+5);$ the sequence is increasing. Since $a_{17} = 33^2 = 1089 \gt 1000$ while $a_{16} = 29 \cdot 33 = 957,$ the greatest term below $1000$ is $a_{16} = 957.$

Therefore $n + a_n = 16 + 957 = 973.$

2004 AIME II Problem 9

Solution:

Problem 9 in Other Years