Geometric Sequences

What makes a sequence geometric

In an arithmetic sequence you reach the next term by adding a fixed number, the common difference $d$. In a geometric sequence you reach it by multiplying by a fixed number. That fixed multiplier is the common ratio, written $r$. To find it, divide any term by the term before it, and in a geometric sequence that quotient is always the same:

Take $3, 6, 12, 24, 48$. Dividing each term by the one before it gives $\tfrac{6}{3} = 2$, $\tfrac{12}{6} = 2$, $\tfrac{24}{12} = 2$, and $\tfrac{48}{24} = 2$. Every ratio is $2$, so the sequence is geometric with $r = 2$. This is the mirror image of the arithmetic test, where you subtract to look for a constant difference. Here the differences are $3, 6, 12, 24$, which are nowhere near constant, so the list is not arithmetic. Multiplying by a constant is a completely different engine from adding one: with $r = 2$ here it makes the terms grow faster and faster.

The common ratio can be almost any number, but the ratio test itself rules out two values. The first term $a_1$ must be nonzero, since finding $r$ starts by dividing by $a_1$. And $r$ itself cannot be $0$: if it were, every term after the first would become $0$, and the very next ratio, $\tfrac{a_3}{a_2}$, would be $\tfrac{0}{0}$, which is undefined. A ratio test that breaks after one step is not a constant ratio. Together these mean no term of a geometric sequence is ever zero: starting away from zero and only ever multiplying by a nonzero $r$ can never land on zero.

Beyond that, $r$ can be almost anything, and its value decides the shape of the sequence. When $r$ is greater than $1$ the terms grow in size, and when $r$ is between $0$ and $1$ they shrink toward zero. When $r$ is negative the terms flip sign back and forth, and when $r$ is exactly $1$ every term is the same.

To test whether a sequence is geometric, check that all the consecutive ratios agree. The sequence $2, 4, 6, 8$ is not geometric: the first ratio is $\tfrac{4}{2} = 2$ but the second is $\tfrac{6}{4} = 1.5$, and a single mismatch disqualifies it. That list is arithmetic instead, since it is the differences that stay constant there.

A formula for the nth term

Multiplying by $r$ over and over reaches any term, but it is slow: to find the $20$th term you would multiply nineteen times. The picture above already shows the shortcut. To land on a later term you apply the ratio a whole number of times from the first term, so you only need to count the multiplications.

The formula has a clean meaning: start at $a_1$ and apply the ratio $n - 1$ times. Just as an arithmetic sequence is pinned down by its first term and common difference, a geometric sequence is pinned down by just two numbers, the first term $a_1$ and the common ratio $r$.

The recursive rule and moving between terms

The formula $a_n = a_1 r^{n-1}$ is the explicit (or closed) form, because it gives a term directly from its position. The same sequence also has a recursive form, which describes one step at a time:

together with a stated first term $a_1$. Read aloud, it is the definition itself: each term is $r$ times the term before it. The recursive rule often matches a real process most honestly (this year’s balance is last year’s balance times the growth factor). But to reach a far-off term with the recursive rule, you must climb through every term in between. The explicit formula is the tool for jumping ahead; the recursive rule is the tool for describing the step.

The multiplying works between any two terms, not only from the first. When $k > m$, getting from the $m$th term to the later $k$th term takes exactly $k - m$ multiplications by $r$, so

The same formula still holds when $k < m$, moving from a later term back to an earlier one, but then “multiply $k - m$ times” stops being the right picture, since $k - m$ is negative. Moving backward divides by $r$ once per step instead, and the negative exponent in $r^{k-m}$ records exactly that: it is $r^{-1}$, which is $\tfrac{1}{r}$, raised to the $m - k$ steps you are moving back. Either way, the formula $a_k = a_m r^{k-m}$ is correct for any two positions $m$ and $k$, and it is the version to reach for when you know two terms and want to recover the ratio or a distant term.

One warning about recovering $r$. In that example the two known terms sat an odd number of positions apart, so $r^3 = 8$ had a single real solution. When the two terms are an even number of positions apart, the equation for $r$ is an even power, and an even power hides a sign. From $a_1 = 3$ and $a_3 = 12$, for instance, you get $r^2 = \tfrac{12}{3} = 4$, which is solved by both $r = 2$ and $r = -2$. The first gives $3, 6, 12, 24$ and the second gives $3, -6, 12, -24$, and both genuinely have $a_3 = 12$. With only those two terms you cannot tell which sequence is meant. To settle it you need one more fact, such as a term that fixes the sign or a statement that the ratio is positive.

Testing membership and solving for n

The explicit formula answers the reverse question too. Instead of asking “what is term number $n$,” you can ask “which term equals a given value,” and you find out by solving for $n$. Because $n$ must be a counting number, this also tells you whether a value appears in the sequence at all.

Matching powers is quickest when the target is a clean multiple of a power of $r$. When it is not, you can still solve $a_1 r^{n-1} = V$ by isolating the power and taking a logarithm, the tool from the logarithms lesson. Dividing gives $r^{n-1} = \tfrac{V}{a_1}$, and a base-$r$ logarithm brings the exponent down to the ground:

This logarithm route applies only when the ratio is positive, with $r > 0$, $r \ne 1$, and $\tfrac{V}{a_1} > 0$, since a base-$r$ logarithm makes sense only then. When $r < 0$ the terms alternate in sign, so there is no base-$r$ logarithm to take. For a negative ratio, test membership by matching powers of $\lvert r \rvert$ and checking that the sign comes out right.

If $\tfrac{V}{a_1}$ is a recognizable power of $r$, the logarithm is a whole number (for $2^{n-1} = 128$ it is $\log_{2}(128) = 7$). If it is not, change of base evaluates it, $\log_{r}(x) = \tfrac{\log(x)}{\log(r)}$, and a non-whole result confirms the value is not a term. The same setup answers a threshold question, such as “which is the first term past a given value.” Solve the inequality for $n$ and keep its direction straight: dividing by $\log(r)$ flips the inequality whenever $0 < r < 1$, since $\log(r)$ is then negative. The bouncing-ball example below works through this carefully.

How the common ratio shapes the sequence

Everything about the long-run behavior of a geometric sequence is decided by the ratio $r$. Four cases cover it, and the positive starting value $16$ keeps the table simple. With a negative starting term, read every row as a statement about size instead: the sign of each term then also depends on the sign of $a_1$.

Two of these deserve a closer look. When $0 < r < 1$, each term is a fraction of the one before it, so the terms fall toward zero and get arbitrarily small. Even so, no term is ever exactly zero, since you only ever multiply a nonzero start by a nonzero ratio. This is exactly the exponential decay you saw with a base between $0$ and $1$. When $r < 0$, multiplying by a negative number flips the sign every step, so the terms swing positive, negative, positive, negative. The sizes of those terms still follow the powers of $\lvert r \rvert$: $r = -2$ still grows in size, $16, -32, 64, -128, \ldots$, while $r = -\tfrac{1}{2}$ still shrinks toward zero, $16, -8, 4, -2, \ldots$. Only the sign is new; the size behavior is the same $r > 1$ versus $0 < r < 1$ story as before. A negative ratio does not mean the sequence decreases; it means it alternates.

Geometric sequences around us

Geometric sequences describe any quantity that is repeatedly scaled by the same factor, the same idea behind the doubling bacteria and the shrinking car value from the opening. A radioactive sample that loses half its atoms in each half-life has $r = \tfrac{1}{2}$, a decay toward zero it never quite reaches. A savings balance left at a fixed annual rate compounded once a year is multiplied by the same growth factor each year, so the yearly balances form a geometric sequence: at a rate of $5$ percent the factor is $r = 1.05$, and the balance after $n - 1$ years is $a_1 r^{n-1}$. That is exactly the compound-interest pattern from the earlier lesson. A ball that rebounds to a fixed fraction of its height on each bounce is the same idea with $0 < r < 1$.

Common mistakes