$5 a_n a_{n+1}=a_n - a_{n+1}$

Calculus Level 2

The sequence $\{a_n\}$ satisfies $a_1 = 1$ and $5 a_n a_{n+1}=a_n - a_{n+1} \text{ for } n \geq 1.$ What is the value of $a_{11}?$

\frac{1}{50}

\frac{1}{49}

\frac{1}{48}

\frac{1}{51}

1 solution

Anish Puthuraya
Mar 6, 2014

$5a_na_{n+1}=a_n-a_{n+1}$

$\Rightarrow 5a_na_{n+1}+a_{n+1} = a_n$

$\Rightarrow a_{n+1} = \frac{a_n}{5a_n+1}$

$\Rightarrow a_{n+1} = \frac{1}{5+\frac{1}{a_n}}$

Substituting a few values,

$a_2 = \frac{1}{6}, a_3 = \frac{1}{11}, a_4 = \frac{1}{16}$

We see that the denominator is an Arithmetic Progression.

Hence,
The series would go on like this : $1,\frac{1}{6},\frac{1}{11},\frac{1}{16}\ldots$

We find $a_{11}$ by simple observation as,

$a_{11} = \frac{1}{51}$

We could also do it this way: $\begin{aligned} &\dfrac 1 {a_{n+1}} - \dfrac 1 {a_n} = 5 \\ \implies& \sum_1^{10} \dfrac 1 {a_{n+1}} - \dfrac 1 {a_n} = 5 = 50 \\ \implies& \dfrac 1 {a_{11}} - \dfrac 1 {a_1} = 50 \\ \implies & \dfrac 1 {a_{11}} = 51 \end{aligned}$

I had almost solved it but in the last step, instead of writing $51$ , I stupidly wrote $49$ .

Parth Thakkar - 7 years, 3 months ago

Thats actually a better one...I have been struggling with recurrence relations problems.. Think you can help me please? Is there any simpler way to solve any recurrence relation?

Anish Puthuraya - 7 years, 3 months ago

Depends upon the relation you're talking about! Some are easy as this one, and some are really tough (at least for me). For example, right now, I am struggling with this one:

Given that $a_{n+1} = 1 + a_1a_2\ldots a_n$ and $a_1 = 1$ , we're to find $\sum_1^\infty \dfrac 1 {a_i}$ .

I have made some progress, but am stuck with a simpler (so it seems!) recurrence relation.

This link seems really good for recurrence relations:

Parth Thakkar - 7 years, 3 months ago

Also, for such relations, my first attempt is to see if we can bring in telescopic cancellations. Like this problem

Parth Thakkar - 7 years, 3 months ago

ya right,after solving few values we can see that last digit of even no of a's are 6 in denominator and last digit of odd no is 1 in denominator and a11, where 11 is odd so last digit will be 1 in denominator so 1/51 is write answer.............

Ragini Maurya - 7 years, 3 months ago

5 a n a n + 1 = a n − a n + 1 5 a_n a_{n+1}=a_n - a_{n+1} 5 a n ​ a n + 1 ​ = a n ​ − a n + 1 ​

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$5 a_n a_{n+1}=a_n - a_{n+1}$