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

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

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

Pre College Mathematics

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

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

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

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

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

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 $p+q=8078$