Geometric Sequences

Definition: A geometric sequence is a sequence for which there exists a constant r such that each term (except for the first term) is r times the previous term. This constant r is called the common ratio of the sequence.

For example,

is a geometric sequence, because each term is 3 times the previous term. Geometric sequences are also known as geometric progressions.

The General Term

Let the first term of a geometric sequence be $a_1$, and the common ratio of the sequence be $r$.

The ratio is always $r = \frac{a_{k+1}}{a_k}$ for any consecutive pair.

For non-consecutive terms $a_m$ and $a_n$,

Example Problem 1

The first term of a geometric sequence of positive numbers is 12, and the fourth term is 24. Find the 10th term of the geometric sequence.

Solution: Let r be the common ratio. We multiply the first term by r three times to get the fourth term, so

Dividing by 12 and taking the cube root of both sides, we find that $r = 2^{1/3}$ is the common ratio of the sequence. The tenth term of the sequence is therefore

The Finite Sum

Proof: $\\$

Let the geometric series have value $S$. Then $S = a_1 + a_1r + a_1r^2 + \cdots + a_1r^{n-1}. \\$ Factoring out $a_1$, multiplying both sides by $(r-1)$, and using the difference of powers factorization yields

Dividing both sides by $r-1$ yields

The Infinite Sum

When $|r| < 1$ the terms shrink to zero and the series converges:

When $|r| \geq 1$ the series diverges, and thus cannot apply this formula.

Proof: If one assumes convergence, Using the terms defined above,

Multiplying both sides by $r$ and adding $a_1$, we find that

Thus, $rS + a_1 = S$, and so $S = \frac{a_1}{1-r}$.

Geometric Mean

Definition: The geometric mean of n positive numbers is the nth root of the product of the n numbers.

Note: For any finite geometric sequence of positive numbers, the geometric mean of the terms in the sequence is equal to the geometric mean of the first term and the last term. (Try Proving this by yourself!)

Proof: Let $n$ be the number of terms of a geometric sequence with first term $a$ and common ratio $r$. The last term of this sequence is $ar^{n-1}$, so the geometric mean of the first and last terms is

The geometric mean of all $n$ terms is

Symmetric Substitution

For problems giving the sum and product of three terms in GP, we write them as $a/r,\ a,\ ar$.

Quantity	Result
Product	$a^3$
Sum	$a(1/r + 1 + r)$
Sum of squares	$a^2(1/r^2 + 1 + r^2)$

The product is shown through $a^3$, so $a$ can be found. Thus, we know that the sum gives a quadratic in $r$.

Concept: When $x \neq 0$ and $x \neq y$, the factorization

is the same as the relationship found by summing the geometric series with first term $x^{n-1}$ and common ratio $y/x$:

More Examples

Example 1: Suppose $x, y,z$ is a geometric sequence with common ratio $r$ and x $\neq$ y. If $x, 2y, 3z$ is an arithmetic sequence, find the value of $r$. (Source: AHSME)

Solution: We can start by reducing the number of variables we have to deal with. From our geometric sequence, we know that $y = rx$ and $z = r^2x$. Now that we can write y and z in terms of r and x, we can write our arithmetic sequence in terms of just x and r. $\\$ So, we know that $x, 2rx,$ and $3r^2x$ are in arithmetic progression, which gives us

Rearranging this equation gives us

Factoring this equation gives

We cannot have x = 0 or r = 1, since both of these give us x = y, which is forbidden in the problem statement. This leaves us with $r = 1/3$ as the only possible value of $r$.

Example 2: The roots of $2x^3 - 19x^2 + kx - 54 = 0$ are in geometric progression for some constant $k$. Find $k$.

Solution: Let the three roots in geometric progression be $\frac{a}{r}, a, ar$ for some values $a$ and $r$. This is a convenient choice because the product of the roots simplifies nicely. $\\$ By Vieta's formulas, the product of the roots equals $\frac{54}{2} = 27$. So,

which gives us $a = 3$. $\\$ By Vieta's formulas, the sum of the roots equals $\frac{19}{2}$. So,

Substituting $a = 3$,

Multiplying through by $2r$,

So $r = \frac{3}{2}$ or $r = \frac{2}{3}$. Both give the same three roots (just in reverse order): $2, 3, \frac{9}{2}$. $\\$ By Vieta's formulas, the sum of the products of the roots taken two at a time equals $\frac{k}{2}$. So,

Therefore $k = \boxed{57}$.

Example 3: The sum of three consecutive terms in a geometric sequence is 39, and the sum of their squares is 741. Find the three terms.

Solution: Let the three consecutive terms be $\frac{a}{r}, a, ar$. We are given two conditions:

Notice that equation $(2)$ is related to the square of equation $(1)$. Squaring equation $(1)$:

Expanding,

Substituting $(2)$ into the left side,

From equation $(1)$, $a\!\left(\frac{1}{r} + 1 + r\right) = 39$, so $\frac{1}{r} + 1 + r = \frac{39}{a}$. Substituting,

Substituting $a = 10$ back into equation $(1)$:

So $r = \frac{5}{2}$ or $r = \frac{2}{5}$. Both give the same three terms in reverse order. $\\$ Taking $r = \frac{2}{5}$: the three terms are $\frac{10}{2/5} = 25,\ 10,\ 4$. $\\$ Taking $r = \frac{5}{2}$: the three terms are $4,\ 10,\ 25$. $\\$ The three terms are $\boxed{4, 10, 25}$.

Example 4: Evaluate the sum $4 - \dfrac{8}{3} + \dfrac{16}{9} - \dfrac{32}{27} + \cdots$

Solution: We first identify this as a geometric series. The first term is $a = 4$. To find the common ratio, we divide the second term by the first:

Since $|r| = \frac{2}{3} < 1$, the series converges. Applying the infinite geometric series formula $S = \dfrac{a}{1 - r}$,

Example 5: Write the following expression as a single polynomial: $\frac{(t^2 + t + 1)(t^{12} + t^9 + t^6 + t^3 + 1)}{t^{10} + t^5 + 1}$ (Source: Mandelbrot)

Solution: We could multiply out the numerator and wade through long division of the product by the denominator. But each factor is a geometric series, so maybe we can use our understanding of geometric series to avoid long division. $\\$ Writing $t^2 + t + 1$ as $1 + t + t^2$, we recognize it as a geometric series with first term $1$ and common ratio $t$, so we have

You might also have recognized this relationship by noting that $t^2 + t + 1$ is a factor of $t^3 - 1$. $\\$ Similarly, the series $1 + t^3 + t^6 + t^9 + t^{12}$ is geometric with first term $1$ and common ratio $t^3$, and the series $1 + t^5 + t^{10}$ is geometric with first term $1$ and common ratio $t^5$. So we have

We see a lot of common factors among our new expressions, so we expect we can do a lot of cancellation:

Cancelling $(t^3 - 1)$ and $(t^{15} - 1)$ from numerator and denominator,

Common Pitfalls

• Mixing up $r^n$ vs. $r^{n-1}$ in the term formula can lead to error.
• Using $r$ from non-consecutive terms without checking consistency.
• Not checking whether $|r| < 1$ before using the infinite formula.

Note: Keep powers as $r^{n-1}$, for convenience and less error (Highly recommended in more complex problems)

Practice Problems