Math Core

Arithmetic sequences change by adding; geometric sequences change by multiplying. They model anything that grows or shrinks by a constant factor, like salaries with a yearly percent raise or a population that doubles. In this lesson you'll review geometric sequences and then derive a formula that adds up many terms at once.

Review: common ratio and the nth term

In $2, 6, 18, 54, \dots$ each term is $3$ times the one before. Dividing any term by the previous term always gives the same number.

Definition

Geometric sequence

A sequence is geometric if the ratio of consecutive terms is constant. That constant $r = \dfrac{a_n}{a_{n-1}}$ is the common ratio ( $r \ne 0$ ).

Starting at $a_1$ , you multiply by $r$ once to get $a_2$ , twice to get $a_3$ , and $n - 1$ times to get $a_n$ :

a_n = a_1 \cdot r^{\,n-1}.

This is an exponential function of $n$ . When $r > 1$ the terms grow; when $0 < r < 1$ they shrink toward $0$ ; when $r < 0$ the signs alternate.

The terms 64, 32, 16, 8, 4, 2 (r = 1/2) lie on the decay curve y = 128 · 0.5^x.Open in grapher →

Worked example: Finding the formula from two terms

In a geometric sequence, $a_3 = 18$ and $a_6 = 486$ . Write an explicit formula.

Going from $a_3$ to $a_6$ multiplies by $r$ three times:

18r^3 = 486 \quad\Rightarrow\quad r^3 = 27 \quad\Rightarrow\quad r = 3.

Back up two steps from $a_3$ : $a_1 = \dfrac{18}{3^2} = 2$ . So $a_n = 2 \cdot 3^{\,n-1}$ .

Check: $a_6 = 2 \cdot 3^5 = 2 \cdot 243 = 486$ . ✓

If the number of steps between the known terms is even, there can be two possible ratios. For example, $r^2 = 9$ allows $r = 3$ or $r = -3$ , and both give valid sequences unless the problem rules one out.

Adding the terms: geometric series

A geometric series is the sum of terms of a geometric sequence. Pairing first and last terms doesn't work here, since the pairs don't have equal sums. Instead, there's a different trick: multiply the sum by $r$ and subtract.

Let $S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}$ . Multiplying by $r$ shifts every term one place:

\begin{aligned} S_n &= a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} \\ r S_n &= \phantom{a_1 + {}} a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} + a_1 r^n \end{aligned}

Subtract the second line from the first. Everything in the middle cancels, leaving

S_n - rS_n = a_1 - a_1 r^n \quad\Rightarrow\quad S_n(1 - r) = a_1(1 - r^n).

Divide by $1 - r$ (allowed as long as $r \ne 1$ ):

Sum of a finite geometric series

For $r \ne 1$ , the sum of the first $n$ terms is

S_n = \frac{a_1\left(1 - r^n\right)}{1 - r}.

If $r = 1$ every term equals $a_1$ , so the sum is just $n a_1$ .

Worked example: Summing the first n terms

Find the sum of the first $8$ terms of $3 + 6 + 12 + 24 + \cdots$ .

Here $a_1 = 3$ , $r = 2$ and $n = 8$ :

S_8 = \frac{3\left(1 - 2^8\right)}{1 - 2} = \frac{3(1 - 256)}{-1} = \frac{3(-255)}{-1} = 765.

Worked example: A negative ratio

Find $5 - 10 + 20 - 40 + \cdots$ through $7$ terms.

Here $a_1 = 5$ and $r = -2$ . Since $(-2)^7 = -128$ ,

S_7 = \frac{5\left(1 - (-2)^7\right)}{1 - (-2)} = \frac{5(1 + 128)}{3} = \frac{645}{3} = 215.

Check by adding: $5 - 10 + 20 - 40 + 80 - 160 + 320 = 215$ . ✓

Common mistake

Keep negative ratios in parentheses. $(-2)^7 = -128$ , but $-2^7$ is read as $-(2^7)$ . That happens to match for odd powers, but for even powers $(-2)^8 = 256$ while $-2^8 = -256$ . Also, $n$ is the number of terms, which is not always the exponent on the last term: the last term is $a_1 r^{n-1}$ .

Series in sigma notation

A sum like $\displaystyle\sum_{k=1}^{n} a\cdot r^{k-1}$ is a geometric series with first term $a$ , ratio $r$ and $n$ terms. If the exponent is written differently, just plug in the lower limit to find the first term.

Worked example: A real-world series

Maya starts a job with a salary of $40,000 and gets a $3\%$ raise every year. What are her total earnings over her first $10$ years, to the nearest dollar?

Each year's salary is $1.03$ times the previous one, so the salaries form a geometric sequence with $a_1 = 40{,}000$ and $r = 1.03$ . The total is

S_{10} = \sum_{k=1}^{10} 40{,}000(1.03)^{k-1} = \frac{40{,}000\left(1 - 1.03^{10}\right)}{1 - 1.03} \approx \frac{40{,}000(-0.343916)}{-0.03} \approx 458{,}555.

She earns about $458,555 in total. Notice how much quicker this is than computing ten salaries and adding them.

Tip

When $r > 1$ , you can flip signs in the numerator and denominator to avoid negatives: $S_n = \dfrac{a_1\left(r^n - 1\right)}{r - 1}$ . It's the same formula.