Sequences That Grow or Shrink by Multiplying
Look carefully at these two sequences:
\[ 2, 4, 8, 16, 32, \dots \]
\[ 81, 27, 9, 3, \dots \]
The first sequence is increasing and the second is decreasing. A natural first idea is to subtract. In the first sequence, \( 4 - 2 = 2 \), \( 8 - 4 = 4 \) and \( 16 - 8 = 8 \). The differences keep changing, so these numbers are not made by adding the same amount each time. The second sequence behaves the same way: \( 27 - 81 = -54 \) but \( 9 - 27 = -18 \).
Now divide each term by the term just before it. In the first sequence, \( 4 \div 2 = 2 \), \( 8 \div 4 = 2 \), \( 16 \div 8 = 2 \) and \( 32 \div 16 = 2 \). The answer is always 2, so every term is twice the one before it. In the second sequence, \( 27 \div 81 = \frac{1}{3} \), \( 9 \div 27 = \frac{1}{3} \) and \( 3 \div 9 = \frac{1}{3} \). Every term is one third of the one before it.
So the two sequences share one feature. Each term is made by multiplying the previous term by a fixed number. In the first the fixed number is 2, which makes the terms grow. In the second it is \( \frac{1}{3} \), which makes them shrink. Sequences built this way have their own name.
Geometric Sequence (Geometric Progression)
A geometric sequence, also called a geometric progression (GP), is a sequence of numbers that increases or decreases by a constant ratio. Each term after the first is obtained by multiplying the term before it by the same fixed number.
Here are three more geometric sequences. Notice what each one is multiplied by at every step.
- \( 3, 12, 48, 192, \dots \) Each term is multiplied by 4, so the terms grow quickly.
- \( 1000, 100, 10, 1, \dots \) Each term is multiplied by \( \frac{1}{10} \), so the terms get smaller.
- \( 1, -3, 9, -27, \dots \) Each term is multiplied by \( -3 \), so the signs switch between positive and negative.
You meet geometric sequences outside the classroom too. Fold a sheet of paper in half and it has 2 layers. Fold it again and it has 4, then 8, then 16. The number of layers is the geometric sequence \( 1, 2, 4, 8, 16, \dots \) because every fold doubles what was there before.
The Common Ratio
The fixed number by which each term of a geometric sequence is multiplied to get the next term is called the common ratio. It is denoted by r, and it is found by dividing any term by the term just before it.
\[ r = \frac{\text{a term}}{\text{preceding term}} \]
If we call the terms \( t_1, t_2, t_3, \dots \), then the ratio is the same whichever pair of neighbouring terms we choose:
\[ r = \frac{t_2}{t_1} = \frac{t_3}{t_2} = \frac{t_4}{t_3} \]
For the sequence \( 2, 4, 8, 16, \dots \) the common ratio is \( r = 2 \). For \( 81, 27, 9, 3, \dots \) it is \( r = \frac{1}{3} \). For \( 1, -3, 9, -27, \dots \) it is \( r = \frac{-3}{1} = -3 \). A common ratio can be a whole number, a fraction or a negative number.
This gives us a test. To decide whether a sequence is geometric, work out the ratio of every pair of neighbouring terms. If all the ratios are equal, the sequence is geometric. If even one ratio is different, it is not. Take \( 2, 6, 18, 50, \dots \). Here \( \frac{6}{2} = 3 \) and \( \frac{18}{6} = 3 \), but \( \frac{50}{18} \) is not 3, so this sequence is not geometric. For it to be geometric, the fourth term would have to be \( 18 \times 3 = 54 \).
Always divide a term by the term BEFORE it, never by the term after it. In \( 81, 27, 9, \dots \), dividing 81 by 27 gives 3, which is wrong. The sequence is shrinking, and the common ratio is \( 27 \div 81 = \frac{1}{3} \).
General Term (nth Term) of a Geometric Sequence
Let the first term of a geometric sequence be a and its common ratio be r. To get the second term we multiply a by r, giving ar. To get the third term we multiply ar by r, giving \( ar^2 \). Multiplying once more gives the fourth term, \( ar^3 \). So every geometric sequence can be written as:
\[ a,\; ar,\; ar^2,\; ar^3,\; \dots \]
Let \( t_1, t_2, t_3, \dots, t_n \) be the first, second, third and nth terms. Watch the power of r in each term and compare it with the position of the term.
| Term | Written with a and r | Power of r |
| First term \( t_1 \) | \( a = ar^{1-1} \) | 0 |
| Second term \( t_2 \) | \( ar = ar^{2-1} \) | 1 |
| Third term \( t_3 \) | \( ar^2 = ar^{3-1} \) | 2 |
| Fourth term \( t_4 \) | \( ar^3 = ar^{4-1} \) | 3 |
| nth term \( t_n \) | \( ar^{n-1} \) | \( n-1 \) |
The power of r is always one less than the position of the term. There is a simple reason. The first term has not been multiplied by r at all. Each move to the next term multiplies by r one more time. To travel from the first term to the nth term you make \( n-1 \) moves, so r is multiplied in \( n-1 \) times.
The general term, or nth term, of a geometric sequence is the formula that gives any term from its position n, the first term a and the common ratio r.
\[ t_n = ar^{n-1} \]
If we know the first term and the common ratio, we can find any term of a geometric progression.
The power is \( n-1 \), not n. Also, in \( ar^{n-1} \) the power belongs to r alone, not to a. If \( t_n = 3 \times 2^{n-1} \), then \( t_5 = 3 \times 2^4 = 3 \times 16 = 48 \). It is not \( 6^4 \).
The formula connects four quantities: the term \( t_n \), the first term a, the common ratio r and the position n. Whenever three of them are known, the fourth can be found. The worked examples below show each kind of question in turn.
Worked Example: Finding a Term
Find the 10th term of the sequence \( 3, 9, 27, \dots \)
Step 1: Checking that the sequence is geometric by dividing each term by the one before it:
\[ \frac{t_2}{t_1} = \frac{9}{3} = 3 \]
\[ \frac{t_3}{t_2} = \frac{27}{9} = 3 \]
Step 2: The ratios are equal, so the sequence is a GP. Reading off its first term and common ratio:
\[ a = 3, \quad r = 3 \]
Step 3: Putting \( n = 10 \) into \( t_n = ar^{n-1} \):
\[ t_{10} = 3 \times 3^{10-1} = 3 \times 3^{9} \]
Step 4: Multiplying powers of the same base by adding their indices, since \( 3 = 3^1 \):
\[ t_{10} = 3^{10} = 59049 \]
Hence the 10th term is 59,049. Writing out ten terms one by one would take much longer, and one slip would spoil every term after it. The formula goes straight to the answer.
Worked Example: Finding the Common Ratio
What is the common ratio of a geometric sequence whose first term is 48 and whose 4th term is 6? Here \( a = 48 \), \( t_4 = 6 \), and r is unknown.
Step 1: Writing the formula for the 4th term:
\[ t_4 = ar^{4-1} = ar^3 \]
Step 2: Substituting \( t_4 = 6 \) and \( a = 48 \):
\[ 6 = 48 \times r^3 \]
Step 3: Dividing both sides by 48 to leave \( r^3 \) alone:
\[ r^3 = \frac{6}{48} = \frac{1}{8} \]
Step 4: Writing \( \frac{1}{8} \) as a cube, because \( 2 \times 2 \times 2 = 8 \):
\[ r^3 = \left(\frac{1}{2}\right)^3 \]
Step 5: Two equal cubes have equal bases, so:
\[ r = \frac{1}{2} \]
Hence the common ratio is \( \frac{1}{2} \). A quick check: halving 48 three times gives 24, then 12, then 6, which is exactly the 4th term.
Worked Example: When Two Terms Are Known
The 3rd term and the 7th term of a GP are 8 and 128. Find the common ratio and the first term, the geometric sequence, and the 15th term.
Here both a and r are unknown. Each given term produces one equation, and dividing one equation by the other removes a, leaving an equation in r alone.
Step 1: Using \( t_n = ar^{n-1} \) for the 3rd term, where \( t_3 = 8 \):
\[ ar^{2} = 8 \quad \text{(i)} \]
Step 2: Using the same formula for the 7th term, where \( t_7 = 128 \):
\[ ar^{6} = 128 \quad \text{(ii)} \]
Step 3: Dividing equation (ii) by equation (i), so that a cancels:
\[ \frac{ar^6}{ar^2} = \frac{128}{8} \]
Step 4: Simplifying both sides, since \( r^6 \div r^2 = r^4 \):
\[ r^4 = 16 = (\pm 2)^4 \]
Step 5: Both 2 and \( -2 \) have a fourth power of 16, so:
\[ r = \pm 2 \]
Step 6: Taking \( r = 2 \), since the GP is in increasing order, and putting it into equation (i):
\[ a \times 2^2 = 8 \]
Step 7: Dividing both sides by 4:
\[ a = \frac{8}{4} = 2 \]
Step 8: Building the sequence from \( a = 2 \), with the second term \( ar = 2 \times 2 = 4 \):
\[ 2, 4, 8, 16, \dots \]
Step 9: Writing the 15th term with the formula:
\[ t_{15} = ar^{14} = 2 \times 2^{14} \]
Step 10: Adding the indices of the equal bases:
\[ t_{15} = 2^{15} = 32768 \]
So the common ratio is 2, the first term is 2, the sequence is \( 2, 4, 8, \dots \) and the 15th term is 32,768. The value \( r = -2 \) also fits the given terms. It gives the same first term, since \( a \times (-2)^2 = 8 \) still means \( a = 2 \), and the sequence \( 2, -4, 8, -16, \dots \). Its 15th term is \( 2 \times (-2)^{14} \), which is again 32,768, because an even power of a negative number is positive.
When an even power of r equals a positive number, as in \( r^4 = 16 \) or \( r^2 = 9 \), there are two answers, one positive and one negative. Write both, then decide which one the question allows. An odd power, such as \( r^3 = \frac{1}{8} \), has only one real answer.
Worked Example: Finding the Number of Terms
Sometimes the first and last terms are given and the question asks how many terms there are. Then n is the unknown, and the last term plays the part of \( t_n \). How many terms are there in \( 2, 6, 18, \dots, 1458 \)?
Step 1: Here \( a = 2 \) and \( r = \frac{6}{2} = 3 \). Setting the last term equal to \( ar^{n-1} \):
\[ 2 \times 3^{n-1} = 1458 \]
Step 2: Dividing both sides by 2:
\[ 3^{n-1} = 729 \]
Step 3: Writing 729 as a power of 3, since \( 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729 \):
\[ 3^{n-1} = 3^{6} \]
Step 4: The bases are equal, so the indices are equal:
\[ n - 1 = 6 \]
Step 5: Adding 1 to both sides:
\[ n = 7 \]
There are 7 terms. Listing them confirms it: 2, 6, 18, 54, 162, 486, 1458.
Worked Example: Three Consecutive Terms
Suppose a, b and c are three consecutive terms of a geometric sequence. The ratio stays the same, so \( \frac{b}{a} = \frac{c}{b} \). Cross multiplying gives a rule worth remembering: the square of the middle term equals the product of the other two.
\[ b^2 = ac \]
This rule finds an unknown inside the terms. Find x if \( x - 2 \), \( x + 1 \) and \( x + 7 \) are consecutive terms of a geometric sequence.
Step 1: Setting the two ratios equal:
\[ \frac{x+1}{x-2} = \frac{x+7}{x+1} \]
Step 2: Cross multiplying:
\[ (x+1)^2 = (x-2)(x+7) \]
Step 3: Expanding both sides:
\[ x^2 + 2x + 1 = x^2 + 5x - 14 \]
Step 4: Subtracting \( x^2 \) from both sides and collecting the x terms on one side:
\[ 3x = 15 \]
Step 5: Dividing both sides by 3:
\[ x = 5 \]
With \( x = 5 \) the three terms are 3, 6 and 12. Each is double the one before, so the common ratio is 2 and the answer checks.
The Formulas at a Glance
| What you want | What to use |
| Common ratio | \( r = \frac{t_2}{t_1} = \frac{t_3}{t_2} \) |
| Any term | \( t_n = ar^{n-1} \) |
| a and r from two terms | Write two equations and divide one by the other |
| Number of terms | Put the last term equal to \( ar^{n-1} \) and solve for n |
| Consecutive terms a, b, c | \( b^2 = ac \) |
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