Geometric Sequences

A geometric sequence changes by a fixed multiplying factor from one term to the next, instead of a fixed added amount. This produces its own version of the position formula and sum formula, both built from powers of the common ratio rather than simple multiples of it.

What you will learn

• Define a geometric sequence by its common ratio
• Find a term using the position formula
• Sum a finite geometric sequence

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Recognizing a geometric sequence

A sequence is geometric when every term is the previous term multiplied by the same fixed number, called the common ratio. In $2,6,18,54,\ldots$, each term is 3 times the one before it, so the common ratio is $r=3$. Checking a sequence for this property means dividing each term by the one before it and confirming every result is identical.

A common ratio between 0 and 1, such as $r=\frac{1}{2}$, produces a decreasing sequence of shrinking positive terms, like $16,8,4,2,\ldots$. A negative common ratio makes the terms alternate in sign. In every case, the defining test is the same: the ratio between consecutive terms must stay constant.

Finding a term directly from its position

Just as an arithmetic sequence has a position formula built from repeated addition, a geometric sequence has one built from repeated multiplication: the $n$th term is $a_n=a_1\times r^{n-1}$. Reaching the $n$th term from the first term requires multiplying by $r$ exactly $n-1$ times, matching the same off-by-one pattern as the arithmetic case.

This formula also runs in reverse: if two terms of a geometric sequence are known, dividing one by the other and taking the appropriate root recovers the common ratio. For instance, if the first term is 5 and the 4th term is 135, then $r^3=\frac{135}{5}=27$, so $r=3$.

Summing a finite geometric sequence

The sum of a finite geometric sequence does not pair off as neatly as an arithmetic sum, but it has its own formula: $S_n=\dfrac{a_1(r^n-1)}{r-1}$, valid whenever $r\neq1$. This formula comes from multiplying the whole sum by $r$ and subtracting the original sum, which cancels every middle term and leaves only a first and a last piece — the same shift-and-subtract idea that the next lesson develops further.

For example, the sum $1+2+4+8+16+32$ has $a_1=1$, $r=2$, and $n=6$ terms, giving $S_6=\dfrac{1(2^6-1)}{2-1}=2^6-1=63$. As with arithmetic sequences, it is essential to count the number of terms $n$ correctly before substituting into the formula.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Using $n$ powers of $r$ instead of $n-1$ powers when finding a term from its position.
• Confusing the geometric sum formula with the arithmetic sum formula and pairing terms that do not actually pair evenly.
• Forgetting to take a root when solving for the common ratio from two non-adjacent terms.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• A geometric sequence multiplies by the same common ratio $r$ to get from each term to the next.
• The $n$th term is $a_n=a_1\times r^{n-1}$, using $n-1$ multiplications from the first term.
• The sum of $n$ terms is $S_n=\dfrac{a_1(r^n-1)}{r-1}$ for $r\neq1$, derived by multiplying the sum by $r$ and subtracting to cancel the middle terms.