2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 15 Medium

A sequence of numbers is defined recursively by a_1=1, a_2=\frac{3}{7}, and a_n=\frac{a_{n-2}\cdot a_{n-1}}{2a_{n-2}-a_{n-1}} for all n>3. Then a_{2019} can be written as \frac{p}{q}, where p and q are relatively prime positive integers. What is p+q? (2019 AMC 10A Problem, Question#15)

  • A.

    2020

  • B.

    4039

  • C.

    6057

  • D.

    6061

  • E.

    8078

Answer:E

Using the recursive formula, we find a_3=\frac{3}{11}, a_4=\frac{3}{15}, and so on. It appears that a_n=\frac{3}{4n-1}, for all n. Setting n=2019, we find a_{2019}=\frac{3}{8075}, so the answer is 8078.

To prove this formula, we use induction. We are given that a_1=1 and a_2=\frac{3}{7}, which satisfy our formula. Now assume the formula holds true for all n\leqslant m for some positive integer m. By our assumption, a_{m-1}=\frac{3}{4m-5} and a_m=\frac{3}{4m-1}. Using the recursive formula,

a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}

=\frac{(\frac{3}{4m-5}\cdot\frac{3}{4m-1})(4m-5)(4m-1)}{(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1})(4m-5)(4m-1)}

=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},

so our induction is complete.

Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by b_n=\frac{1}{a_n}.

We have , \frac{1}{a_n}=\frac{2a_{n-2}-a_{n-1}}{a_{n-2}\cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}} so in other words, b_n=2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\cdots.

By recursively following this pattern, we can see that b_{n}=(n-1)\cdot b_2-(n-2)\cdot b_1.

By plugging in 2019, we thus find b_{2019}=2018\cdot\frac{7}{3}-2017=\frac{8075}{3} . Since the numerator and the denominator are relatively prime, the answer is 8078.

It seems reasonable to transform the equation into something else. Let a_n=x, a_{n-1}=y, and a_{n-2}=z. Therefore, we have x=\frac{zy}{2z-y}2xz-xy=zy2xz=y(x+z)y=\frac{2xz}{x+z} Thus, y is the harmonic mean of x and z. This implies a_n is a harmonic sequence or equivalently b_n=\frac{1}{a_n} is arithmetic. Now, we have b_1=1, b_2=\frac{7}{3}, b_3=\frac{11}{3}, and so on. Since the common difference is \frac{4}{3}, we can express b_n explicitly as b_n=\frac{4}{3}{n-1}+1. This gives b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3} which implies a_{2019}=\frac{3}{8075}=\frac{p}{q}, p+q=8078.

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