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Geometric Sequences

A geometric sequence is a list of numbers made by multiplying by the same number over and over. That repeated multiplier is called the common ratio, rr. The sequence 2,6,18,54,…2, 6, 18, 54, \ldots is geometric because every term is 33 times the one before it.

Compare that with an arithmetic sequence, which adds the same amount each step. Geometric sequences multiply — which is exactly what exponential functions do. In fact, a geometric sequence is just an exponential function evaluated at n=1,2,3,…n = 1, 2, 3, \ldots

Finding the common ratio

Divide any term by the term before it. For 2,6,18,542, 6, 18, 54: 62=3\dfrac{6}{2} = 3, and checking further, 186=3\dfrac{18}{6} = 3 and 5418=3\dfrac{54}{18} = 3. The ratio is the same every time, so r=3r = 3.

If the terms are shrinking, rr is a fraction: the sequence 80,40,20,1080, 40, 20, 10 has r=12r = \dfrac{1}{2}. If dividing consecutive terms gives different answers, the sequence is not geometric.

The rule for the nnth term

Every geometric sequence follows an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}, where a1a_1 is the first term and rr is the common ratio. The exponent is n−1n - 1, not nn — the first term has been multiplied by rr zero times, the second term once, and so on.

This rule lets you jump straight to any term. Want the tenth term without listing nine others? Plug in n=10n = 10 and evaluate.

Why the exponent is n−1n - 1

Check it against the first term: at n=1n = 1 the rule gives a1⋅r0=a1a_1 \cdot r^0 = a_1, which is correct. If the exponent were nn, the rule would multiply by one extra factor of rr and every term would be wrong. Testing n=1n = 1 takes two seconds and catches this mistake every time.

Worked examples

Example 1: find rr and write the rule

Find the common ratio and the rule for 2,6,18,54,…2, 6, 18, 54, \ldots

Divide consecutive terms62=3\dfrac{6}{2} = 3
Verify with another pair186=3\dfrac{18}{6} = 3
The first term is a1=2a_1 = 2, so write the rulean=2⋅3n−1a_n = 2 \cdot 3^{n-1}

Answer: r=3r = 3; an=2⋅3n−1a_n = 2 \cdot 3^{n-1}

Example 2: find a far-away term

Using an=2⋅3n−1a_n = 2 \cdot 3^{n-1}, find the 6th term.

Substitute n=6n = 6a6=2⋅35a_6 = 2 \cdot 3^{5}
Evaluate the power35=2433^5 = 243
Multiplya6=486a_6 = 486

Answer: a6=486a_6 = 486

Example 3: from a1a_1 and rr

A geometric sequence has a1=5a_1 = 5 and r=2r = 2. Find a4a_4.

Write the rulean=5⋅2n−1a_n = 5 \cdot 2^{n-1}
Substitute n=4n = 4a4=5⋅23a_4 = 5 \cdot 2^{3}
Evaluatea4=5⋅8=40a_4 = 5 \cdot 8 = 40

Answer: a4=40a_4 = 40

Try one yourself

Common questions

How do I tell geometric from arithmetic?

Check what stays constant between terms. Same difference each step (add 44, add 44, …) means arithmetic. Same ratio each step (times 33, times 33, …) means geometric. The sequence 1,4,16,641, 4, 16, 64 multiplies by 44, so it is geometric.

Can the common ratio be a fraction?

Yes. A shrinking sequence like 80,40,20,1080, 40, 20, 10 has r=12r = \dfrac{1}{2}. Each term is half the one before — still geometric, just decaying instead of growing.

Why is the exponent n−1n - 1 instead of nn?

Because the first term has not been multiplied by rr yet. By the nnth term, the ratio has been applied n−1n - 1 times. Test with n=1n = 1: the rule must return a1a_1 itself.

How are geometric sequences related to exponential functions?

They are the same idea with a restricted input. The rule an=a1⋅rn−1a_n = a_1 \cdot r^{n-1} is an exponential function whose inputs are the positive integers 1,2,3,…1, 2, 3, \ldots instead of all real numbers.

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