Recursion Problem | AMC 10A, 2019| Problem No 15

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Try this beautiful Problem on Algebra based on Recursion from AMC 10 A, 2019. You may use sequential hints to solve the problem.

Recursion- AMC-10A, 2019- Problem 15

A sequence of numbers is defined recursively by $a_{1}=1, a_{2}=\frac{3}{7},$ and $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

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 $p+q ?$

$2020$
$4039$
$6057$
$6061$
$8078$

Key Concepts

Algebra

Recursive formula

Suggested Book | Source | Answer

Source of the problem

AMC-10A, 2019 Problem-15

Check the answer here, but try the problem first

$8078$

Try with Hints

First Hint

The given expression is $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$ and given that $a_{1}=1, a_{2}=\frac{3}{7}$. we have to find out $a_{2019}$?

at first we may use recursive formula we can find out $a_3$ , $a_4$ with the help of $a_1$, $a_2$. later we can find out $a_n$

Now can you finish the problem?

Second Hint

Given that $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

Now $n=3$ then $a_{3}=\frac{a_{(3-2)} \cdot a_{(3-1)}}{2 a_{(3-2)}-a_{(3-1)}}$

$\Rightarrow$ $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

$\Rightarrow$ $a_{3}=\frac{1*\frac{3}{7}}{2*1-\frac{3}{7}}$

$\Rightarrow$ $a_{3}=\frac{3}{7}$

Similarly if we put $n=4$ we get $a_4=\frac{3}{15}$ (where $a_{1}=1, a_{2}=\frac{3}{7}$,$a_3=\frac{3}{7}$)

Continue this way we $a_{n}=\frac{3}{4 n-1}$

So can you find out the value of $a_{2019}$?

Now Can you finish the Problem?

Third Hint

Now $a_{n}=\frac{3}{4 n-1}$

Put $n=2019$

$a_{2019}=\frac{3}{8075}$ which is the form of $\frac{p}{q}$

Therefore $p+q=8078$

Recursion Problem | AMC 10A, 2019| Problem No 15