HMMT 二月 2019 · ALGNT 赛 · 第 5 题
HMMT February 2019 — ALGNT Round — Problem 5
题目详情
- Let a , a , . . . be an arithmetic sequence and b , b , . . . be a geometric sequence. Suppose that a b = 20, 1 2 1 2 1 1 a b = 19, and a b = 14. Find the greatest possible value of a b . 2 2 3 3 4 4
解析
- Let a , a , . . . be an arithmetic sequence and b , b , . . . be a geometric sequence. Suppose that a b = 20, 1 2 1 2 1 1 a b = 19, and a b = 14. Find the greatest possible value of a b . 2 2 3 3 4 4 Proposed by: Michael Tang 37 Answer: 4 We present two solutions: the first more algebraic and computational, the second more conceptual. Solution 1. Let { a } have common difference d and { b } have common ratio d ; for brevity, let a = a n n 1 2 and b = b . Then we have the equations ab = 20, ( a + d ) br = 19, and ( a + 2 d ) br = 14, and we want 1 3 to maximize ( a + 3 d ) br . The equation ( a + d ) br = 19 expands as abr + dbr = 19, or 20 r + bdr = 19 since ab = 20. Similarly, 2 2 2 (20 + 2 bd ) r = 14, or 10 r + bdr = 7. Multiplying the first equation by r and subtracting the second, we get 2 10 r = 19 r − 7 = ⇒ (5 r − 7)(2 r − 1) = 0 , 7 1 so either r = or r = . 5 2 19 − 20 r 19 For each value of r , we have bd = = − 20, so r r ( ) 57 3 3 3 2 ( a + 3 d ) br = (20 + 3 bd ) r = − 40 r = r (57 − 40 r ) . r 37 1 The greater value of this expression is , achieved when r = . 4 2 Solution 2. The key is to find a (linear) recurrence relation that the sequence c = a b satisfies. n n n n n Some knowledge of theory helps here: c is of the form snr + tr for some constants r, s, t , so { c } n n 2 2 2 satisfies a linear recurrence relation with characteristic polynomial ( x − r ) = x − 2 rx + r . That is, 2 c = 2 rc − r c n n − 1 n − 2 for some constant r . 7 2 Taking n = 3, we get 14 = 2 r · 19 − r · 20, which factors as (5 r − 7)(2 r − 1) = 0, so either r = or 5 1 r = . Then 2 2 2 c = 2 rc − r c = 28 r − 19 r . 4 3 2 14 1 14 This expression is maximized at r = , and strictly decreases on either side. Since is closer to 19 2 19 7 1 19 37 than , we should choose r = , giving the answer c = 14 − = . 4 5 2 4 4