Recurrence relation in series with parameter $k$

Question

I was given a recurrence relation of this series: $$a_{n+1} = k -a_n$$ $S_n$ is the sum of $n$ first terms of the series. So I was given that $$S_{101} = 353$$ $$S_{199} = 696$$ With this information I need to find $a_1$ and $k$.

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I found that if $$a_{n+1} = k -a_n$$ Then \begin{align} \ a_{n+2} & = k - a_{n+1} \\ & = k - (k-a_n) \\ \end{align} So $$a_{n+2} = a_n$$

I thought that I could find sums by thinking that the series above is an arithmetic progression, $d = 0$ but it didn't help. Later, I thought that it would be a geometric series, the ratio being $r=1$.

The sum of geometric series would be: $$S_n = \frac{a_1(r^n-1)}{r-1}$$ Thus, $r$ can't be $1$.

Any help?

Answer 1

You have reached the conclusion that $a_n=a_{n+2}$.

Now all you have left is to solve the following system of two equations in two variables:

• $51a_1+50a_2=353$
• $100a_1+99a_2=696$

The solution is $a_1=3$ and $a_2=4$.

And then of course, $k=a_1+a_2=7$.

Answer 2

Let's just unroll the recurrence relation.

$ a_2 = k - a_1 $

$ a_3 = k - a_2 = k - (k - a_1) = a_1 $

$ a_4 = k - a_3 = k - a_1 = a_2 $

$ \cdots $

What do you find?

All odd terms in the sequence is simply $ a_1 $

All even terms in the sequence is simply $ a_2 $

The rest is trivial.

$ \sum\limits_{i = 1}^{101}a_i = \sum\limits_{i = 1}^{51}a_{2i - 1} + \sum\limits_{i = 1}^{50}a_{2i} = 51a_1 + 50a_2 = 353 $

$ \sum\limits_{i = 1}^{199}a_i = \sum\limits_{i = 1}^{100}a_{2i - 1} + \sum\limits_{i = 1}^{99}a_{2i} = 100a_1 + 99a_2 = 696 $

Solving, we get $ a_1 = 3 $ and $ a_2 = 4 $, so $ a_2 = k - a_1 \implies 4 = k - 3 \implies k = 7 $