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A geometric sequence, also called a geometric series or geometric progression, is a set of numbers formed by multiplying each previous number in the set by a constant. The factor by which each successive term is multiplied is called the common ratio because it is common to all of the terms in the set. The common ratio cannot equal $$0$$ $$\left(r\ne 0\right)$$.
The standard form of geometric sequences can be expressed as:
$$a,a·r,a·r^2,a·r^3,a·r^4...$$ in which:

• $$a$$ represents the first term and is sometimes written as $$a_1$$.
• $$r$$ represents the common ratio.

Example: if the first term of the sequence is $$1$$ and the common ratio is $$3$$, then each successive term can be obtained by multiplying the previous term by 3, and the sequence will look like this:
$$1,3,9,27,81...$$
which can also be written as:
$$1,1·3,1·3^2,1·3^3,1·3^4...$$

Formulas
Finding any term ($$a_n$$) in a geometric sequence:
$$a_n=a·r^{\left(n-1\right)}$$

• $$a$$ represents the first term.
• $$n$$ represents the position of a term in the sequence. A sequence with $$n$$ number of terms, for example, would be written as:
  $$a,a·r,a·r^2,a·r^3,a·r^4...a·r^{\left(n-1\right)}$$ in which the last term is raised to the power of $$n-1$$ (because the first term is raised to the power of $$0$$).
• $$r$$ represents the common ratio.

Example: To find the next term in $$1,3,9,27,81...$$ which would be the 6th term, we would plug the following into the general term formula, $$a_n=a·r^{\left(n-1\right)}$$:
$$a$$ (first term)$$=1$$
$$r$$ (common ratio)$$=3$$
$$n$$ (term number)$$=6$$.

This would give us $$a_6=1·3^{\left(6-1\right)}$$, which we could solve to get $$a_6=243$$. So, our sequence would be: $$1,3,9,27,81,243...$$

Finding the sum of all the terms in a geometric sequence:
$$s=a\left(\left(1-r^n\right)/\left(1-r\right)\right)$$

• $$s$$ is the sum of the terms in the sequence.
• $$a$$ represents the first term.
• $$n$$ represents the position of a term in the sequence.
• $$r$$ represents the common ratio.

Example: To find the sum of $$1,3,9,27,81$$ we plug the following into the sum formula, $$s=a\left(\left(1-r^n\right)/\left(1-r\right)\right)$$:
$$a$$ (first term)$$=1$$
$$r$$ (common ration)$$=3$$
$$n$$ (total number of terms)$$=5$$.

This would give us $$s=1\left(\left(1-35\right)/\left(1-3\right)\right)$$, which we could solve to get $$s=121$$.