2019 AMC 10A Problem 15

Question

A sequence of numbers is defined recursively by

a_1 = 1,

a_2 = \frac{3}{7},

and

a_n=\dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}

for all

n \geq 3.

Then

a_{2019}

can be written as

\frac{p}{q},

where

p

and

q

are relatively prime positive integers. What is

p+q ?

2020

4039

6057

6061

8078

Take reciprocals of the recurrence

The reciprocal sequence becomes arithmetic

Taking reciprocals in the recursive formula gives

\begin{aligned} \frac1{a_n} &=\frac{2a_{n-2}-a_{n-1}}{a_{n-2}a_{n-1}}\\ &=\frac2{a_{n-1}}-\frac1{a_{n-2}}. \end{aligned}

This means that

\dfrac{1}{a_n} - \dfrac{1}{a_{n - 1}} = \dfrac{1}{a_{n - 1}} - \dfrac{1}{a_{n - 2}},

which tells us that

\left\{\dfrac{1}{a_n}\right\}

is an arithmetic sequence.

Using

a_1

and

a_2,

we get that the common difference is

\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.

Therefore

\frac1{a_{2019}}=1+2018\cdot\frac43=\frac{8075}{3},

so

a_{2019}=\frac3{8075}.

p + q

is therefore

8075 + 3 = 8078.

Thus, E is the correct answer.

Problem 15 in other years