Geometric Sequences

One Fixed Multiplier Between Consecutive Terms

An arithmetic sequence changes by adding the same amount at every step. Another common pattern is to multiply by the same number each time. For instance, the terms $3, -6, 12, -24,\ldots$ are obtained by multiplying each term by $-2$. The multiplier, rather than the difference between terms, determines the pattern.

Definition: A real sequence $(a_n)_{n=0}^{\infty}$ is a geometric sequence if there is a real number $r$ such that $$ a_{n+1}=r a_n $$ for every $n\in\mathbb{N}_0$. The number $r$ is called a common ratio.

The definition uses multiplication, so it remains meaningful when a term is zero. If $a_n\neq0$, the equation can be written as $r=a_{n+1}/a_n$. But division is not part of the definition: if $a_n=0$, the quotient is undefined. In particular, if $a_0=0$, the defining equation forces every term to be zero, and that zero sequence satisfies the definition for every real $r$. Thus its common ratio is not unique.

When $a_0\neq0$, the common ratio is unique: it must equal $a_1/a_0$. A sequence that begins with a nonzero term can never reach zero and later become nonzero, since repeated multiplication by a fixed ratio either keeps all terms nonzero or, when the ratio is zero, makes every term after $a_0$ zero.

The Explicit Formula

Theorem: A sequence $(a_n)_{n=0}^{\infty}$ is geometric with common ratio $r$ if and only if $$ a_n=a_0r^n $$ for every $n\in\mathbb{N}_0$. Here $r^0=1$, including when $r=0$.

Proof. Suppose first that $(a_n)$ is geometric with common ratio $r$. We prove by induction that $a_n=a_0r^n$ for every $n\in\mathbb{N}_0$. At $n=0$, $a_0r^0=a_0$, so the formula holds. Suppose it holds at some $n\in\mathbb{N}_0$. The defining relation gives

a_{n+1}=r a_n=r(a_0r^n)=a_0r^{n+1}.

This proves the induction step, so the formula holds for every nonnegative integer $n$.

Conversely, suppose $a_n=a_0r^n$ for every $n\in\mathbb{N}_0$. Then, for every such $n$,

a_{n+1}=a_0r^{n+1}=r(a_0r^n)=r a_n.

Therefore the sequence is geometric with common ratio $r$. The formula also shows that $a_0$ and $r$ determine every term. If $a_0\neq0$, then $r=a_1/a_0$, so the ratio is unique. If $a_0=0$, the formula gives the zero sequence for every $r$, as described above. $\square$

For any nonnegative index $m$ and any $k\in\mathbb{N}_0$, the formula also gives $a_{m+k}=a_mr^k$. This describes the terms relative to a later starting point. It does not require dividing by $a_m$, so it remains valid if that term is zero.

Worked Example: Checking a Formula and Finding the Ratio

Let $p_n=7\left(-\frac{1}{3}\right)^n$ for $n\in\mathbb{N}_0$. The explicit formula has $p_0=7$ and candidate ratio $r=-\frac13$. Directly, for every $n$,

p_{n+1} =7\left(-\frac{1}{3}\right)^{n+1} =-\frac{1}{3}\left(7\left(-\frac{1}{3}\right)^n\right) =-\frac{1}{3}p_n.

Thus the sequence is geometric with common ratio $-\frac13$. Its first terms are $p_0=7$, $p_1=-\frac73$, and $p_2=\frac79$. The multiplications check: $-\frac13(7)=-\frac73$, and $-\frac13(-\frac73)=\frac79$.

Finding the Ratio from Given Terms

If two terms at different indices are known, their relationship can determine the ratio, provided the sequence is already known to be geometric. The number of steps between the indices determines which power of the ratio relates the terms.

Proposition: If $(a_n)$ is geometric with common ratio $r$, then for nonnegative integers $m$ and $k$, $$ a_{m+k}=a_mr^k. $$ In particular, if $n>m$, then $a_n=a_mr^{n-m}$.

This follows from the explicit formula: $a_{m+k}=a_0r^{m+k}=(a_0r^m)r^k=a_mr^k$. If $a_m\neq0$, the relation between terms at indices $m$ and $n$ may be used to find a candidate ratio. For example, it gives $r^{n-m}=a_n/a_m$. A solution must still be checked against the given information and the real values allowed for $r$.

Worked Example: Determining the Ratio from Two Terms

Suppose $(q_n)$ is geometric, $q_2=12$, and $q_5=-96$. The indices differ by three, so

q_5=q_2r^3, \qquad -96=12r^3, \qquad r^3=-8.

The real solution is $r=-2$, since $(-2)^3=-8$. Using $q_2=q_0r^2$, we find $12=q_0(-2)^2=4q_0$, so $q_0=3$. The resulting formula is $q_n=3(-2)^n$. Substitution verifies both given terms: $q_2=3(-2)^2=3(4)=12$, and $q_5=3(-2)^5=3(-32)=-96$.

The nonzero-term condition matters when using the quotient $a_n/a_m$. If $a_m=0$, division by $a_m$ is not valid. The relation $a_n=a_mr^{n-m}$ is still valid, but it then says $a_n=0$ and may provide no information about $r$.

Worked Example: Why a Zero Term Needs Care

Let $t_0=5$ and $t_1=0$, and suppose the sequence is geometric. From $t_1=rt_0$, we have $0=5r$, so $r=0$. The explicit formula gives $t_n=5(0)^n$: $t_0=5$, while $t_n=0$ for every $n\geq1$. The ratio is determined here because $t_0\neq0$.

In contrast, if $t_0=0$ and the sequence is geometric, then $t_n=0$ for every $n$, regardless of the value of $r$. For example, both $r=2$ and $r=-5$ yield $t_n=0$ at every index. It would be incorrect to infer a unique ratio by dividing $t_1$ by $t_0$.

How the Ratio Affects the Direction of Change

The ratio does not always make a sequence increase or decrease. A useful way to decide is to compare consecutive terms by subtracting: the difference factors into the current term and a quantity determined by the ratio.

Theorem: Let $(a_n)$ be geometric with common ratio $r>0$. If $a_0>0$, then the sequence is strictly increasing when $r>1$, strictly decreasing when $0<r<1$, and constant when $r=1$. If $a_0<0$, the increasing and decreasing conclusions are reversed. If $a_0=0$, the sequence is constant.

Proof. If $a_0>0$ and $r>0$, the formula $a_n=a_0r^n$ shows $a_n>0$ for every $n$. For each index,

a_{n+1}-a_n=ra_n-a_n=a_n(r-1).

When $r>1$, both factors on the right are positive, so $a_{n+1}-a_n>0$ at every index and the sequence is strictly increasing. When $0<r<1$, $a_n>0$ and $r-1<0$, so $a_{n+1}-a_n<0$ at every index and the sequence is strictly decreasing. When $r=1$, the difference is zero at every index, so the sequence is constant.

If $a_0<0$ and $r>0$, every $a_n$ is negative. For $r>1$, the product $a_n(r-1)$ is negative, giving strict decrease; for $0<r<1$, both factors are negative, giving a positive difference and strict increase. If $r=1$, the difference is again zero. Finally, if $a_0=0$, the explicit formula gives $a_n=0$ for every $n$, so the sequence is constant. $\square$

A negative ratio has a different effect: if $a_0\neq0$, consecutive terms have opposite signs, because $r^n$ alternates in sign. Such a sequence is not monotone: its first three terms alternate direction, so it cannot be entirely increasing or entirely decreasing. If $r=0$, the sequence consists of $a_0$ followed by zeros; it is constant only when $a_0=0$.

Worked Example: A Negative Starting Term with a Fractional Ratio

Let $v_n=-8\left(\frac14\right)^n$. Here $v_0=-8<0$ and $0<r=\frac14<1$. The theorem predicts that the sequence is strictly increasing. The first terms confirm the direction:

v_0=-8,\qquad v_1=-8\left(\frac14\right)=-2,\qquad v_2=-8\left(\frac{1}{16}\right)=-\frac12.

Indeed, $-8<-2<-\frac12$. The values get closer to zero from below; the fact that their absolute values decrease does not mean that the sequence itself decreases.

A Formula for Finite Geometric Sums

The explicit formula also makes it possible to add a finite block of geometric terms. The sum formula below is new in its general form; the finite sum of powers of two established earlier in the course is one special case.

Theorem: Let $(a_n)$ be geometric with $a_n=a_0r^n$. For every $n\in\mathbb{N}_0$, if $r\neq1$, then $$ \sum_{k=0}^{n}a_k =a_0\frac{1-r^{n+1}}{1-r}. $$ If $r=1$, then $\displaystyle\sum_{k=0}^{n}a_k=(n+1)a_0$.

Proof. Write $S_n=\sum_{k=0}^{n}a_k$. If $r\neq1$, the explicit formula gives

S_n=a_0+a_0r+a_0r^2+\cdots+a_0r^n.

Multiplying this equality by $r$ gives $rS_n=a_0r+a_0r^2+\cdots+a_0r^{n+1}$. Subtract this equality from the expression for $S_n$. All the intermediate terms cancel, leaving

S_n-rS_n=a_0-a_0r^{n+1}.

Thus $(1-r)S_n=a_0(1-r^{n+1})$. Since $r\neq1$, division by $1-r$ is valid, and the stated formula follows. If $r=1$, then $a_k=a_0$ for every $k$, so the sum has $n+1$ equal terms and is $(n+1)a_0$. $\square$

Worked Example: Summing a Geometric Sequence with Negative Ratio

Consider the first five terms of the sequence $w_k=2(-3)^k$, for $k=0,1,2,3,4$. Here $a_0=2$, $r=-3$, and $n=4$. Since $r\neq1$, the finite-sum formula gives

\sum_{k=0}^{4}w_k =2\frac{1-(-3)^5}{1-(-3)} =2\frac{1-(-243)}{4} =2\frac{244}{4} =122.

A direct check gives the terms $2,-6,18,-54,162$. Their sum is $2-6+18-54+162=122$, agreeing with the formula. The alternating signs do not prevent the finite-sum identity from applying.

Using the Definition Rather Than Guessing from a Few Terms

A sequence is geometric only if one fixed multiplier works at every index. When the terms are all nonzero, equal ratios between several displayed consecutive terms can suggest a common ratio, but a few checks do not establish the pattern for every term. For a formula, the reliable test is to verify $a_{n+1}=ra_n$ for all allowed $n$, or to establish the explicit formula $a_n=a_0r^n$.

For instance, the first terms $2,6,18$ suggest a ratio of $3$, but those three values alone say nothing about what comes next unless a rule or a geometric-sequence assumption is supplied. By contrast, if a proposed formula is $a_n=2\cdot3^n$, then substitution shows $a_{n+1}=2\cdot3^{n+1}=3a_n$ for every index, which proves the sequence is geometric.

The main distinctions are between a fixed additive change and a fixed multiplicative change, and between a common ratio and a quotient that may be undefined. The explicit formula translates the recursive rule into a direct expression for any term, while the finite-sum theorem provides a way to add a consecutive block of terms without listing them individually.

Check Your Understanding

Use the definition and the proved results to analyze the following sequences.

For $u_n=6\left(-2\right)^n$, identify $u_0$ and the common ratio. What are $u_1$ and $u_2$?
A geometric sequence has $v_1=4$ and $v_3=36$. What real values of the common ratio are possible, and why?
Explain why a geometric sequence with $a_0=0$ cannot have a unique common ratio.
For $w_n=-5\left(\frac12\right)^n$, is the sequence increasing or decreasing? Justify your answer using the sign of $a_0$ and the size of $r$.
Use the finite geometric sum formula to find $\sum_{k=0}^{3}3\left(\frac12\right)^k$.

What You'll Learn