# Finding Intermediate Values in Arithmetic Mean Sequence

The following article is inspired from https://brilliant.org/daily-problems/average-visual/ of March 17th, 2019.

The Arithmetic Mean Sequence for any 2 numbers, a­1 and a2, is a progressive series of numbers where each number is the Arithmetic Mean of the previous 2 numbers, starting the sequence with a­1 and a2.

## Solving intermediary values in the sequence

Let x be the lower end of the line passing between points x and y, such that varn lies on the line, and y is the Arithmetic Mean of its previous 2 numbers.

Let $var_n | n\ \in\mathbb{N}\ \bigwedge\ n>2$ be the Arithmetic Mean average of the previous 2 values in the list $$x, var_1, var_2, …, y$$

The intermediary values can be found in terms of the first unknown number, var1. This approach requires that the var­1 is computed first by using the only 2 numbers known.

$${var}_1=\frac{2^{\left|vars\right|}y\ -\ Jacobsthal(\left|vars\right|)x}{Jacobsthal(\left|vars\right|\ +\ 1)}$$

Once the first variable is found, the next unknown values can be computed in terms of the known values.

Two approaches are possible to calculate the remaining values, once the first value is known. The easiest way is to use the Arithmetic Mean. That is, to build the sequence one unknown at a time. However, if the list of unknown values is unknown a more direct approach to find the unknown values at position n is to calculate the value in terms of the first unknown value that was found, using the equation

$${var}_n=\frac{Jacobsthal(n\ -\ 1)x\ +\ Jacobsthan(n){var}_1}{2^{n-1}}\ |\ n\ \geq\ 2$$

The above is good, but with more elaboration, the necessity to compute the first unknown value can be removed. With some manipulation to the above process, it is possible to compute each unknown value in terms of the boundary known values. The equation for direct calculation of an unknown value in the sequence is

$${var}_n=\frac{2^{\left|vars\right|}Jacobsthal(n)y\ +\ \left({-1}^nJacobsthal(\left|vars\right|)\ +\ {-1}^{\left|vars\right|}Jacobsthal(n\ -\ 1)\right)x}{Jacobsthal(\left|vars\right|\ +\ 1)2^{n-1}}$$

Using the above equations it is possible to solve intermediate values of a finite arithmetic mean sequence.

Proofs for the above equations can be found in the supplementary paper.