# Product of Geometric Sequences from One

## Theorem

Let $Q_1 = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ and $Q_2 = \sequence {b_j}_{0 \mathop \le j \mathop \le n}$ be geometric sequences of integers of length $n + 1$.

Let $a_0 = b_0 = 1$.

Then the sequence $P = \sequence {p_j}_{0 \mathop \le j \mathop \le n}$ defined as:

$\forall j \in \set {0, \ldots, n}: p_j = b^j a^{n - j}$

is a geometric sequence.

In the words of Euclid:

If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion.

## Proof

By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $Q_1$ is given by:

$a_j = a^j$

and of $Q_1$ is given by:

$b_j = b^j$

Let the terms of $P$ be defined as:

$\forall j \in \left\{{0, 1, \ldots, n}\right\}: p_j = b^j a^{n - j}$

Then from Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence.

$\blacksquare$