# Form of Geometric Sequence of Integers

## Theorem

Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n + 1$ consisting of integers only.

Then the $j$th term of $P$ is given by:

- $a_j = k p^{n - j} q^j$

where:

- the common ratio of $P$ expressed in canonical form is $\dfrac q p$
- $k$ is an integer.

### Corollary

Let $p$ and $q$ be integers.

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

- $a_j = p^j q^{n - j}$

is a geometric sequence whose common ratio is $\dfrac p q$.

## Proof

Let $r$ be the common ratio of $P$.

From Common Ratio in Integer Geometric Sequence is Rational, $r$ is a rational number.

Let $r = \dfrac q p$ be in canonical form.

Thus, by definition:

- $p \perp q$

Let $a$ be the first term of $P$.

Then the sequence $P$ is:

- $P = \paren {a, a \dfrac q p, a \dfrac {q^2} {p^2}, \ldots, a \dfrac {q^n} {p^n} }$

All the elements of $P$ are integers, so, in particular:

- $a \dfrac {q^n} {p^n} \in \Z$

We have that:

- $p \perp q$

From Powers of Coprime Numbers are Coprime:

- $q^n \perp p^n$

and so from Euclid's Lemma:

- $p^n \divides a$

Thus:

- $a = k p^n$

for some $k \in \Z$, and so:

- $P = \paren {k p^n, k q p^{n - 1}, k q^2 p^{n - 2}, \ldots, k q^{n - 1} p, k q^n}$

$\blacksquare$