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Algebra I/Functions & Exponentials/Geometric Sequences
Algebra I Regents in 22 days
Algebra I · Lesson 5

Geometric Sequences

Multiply by the same number every time, and you get a pattern with real power.


In 2020, a single COVID-19 case in a city became dozens within a week, then hundreds, then thousands. Each day, the number of cases wasn't growing by a fixed amount — it was roughly doubling. That kind of growth, where each term is multiplied by the same number to get the next, is called a geometric sequence.

A geometric sequence is a list of numbers where you multiply by the same value each time to move from one term to the next. That fixed multiplier is called the common ratio, usually written rr. To find it, divide any term by the one before it.

Look at this sequence: 3,6,12,24,48,3, 6, 12, 24, 48, \ldots

Each term is double the previous one. So r=2r = 2. Compare that to an arithmetic sequence like 3,6,9,12,15,3, 6, 9, 12, 15, \ldots — there you add 3 each time. Here you multiply by 2. The difference is everything.

To find any term directly, without listing every term before it, use the explicit formula:

a_n = a_1 \cdot r^

Here, ana_n is the term you want, a1a_1 is the first term, rr is the common ratio, and nn is the position of the term. The exponent is n1n - 1, not nn, because the first term requires zero multiplications — you start at a1a_1 and haven't multiplied yet.

Find the 6th term of 3, 6, 12, 24, ...
r=63=2r = \frac{6}{3} = 2
Divide any term by the one before it to find the common ratio.
an=a1rn1a_n = a_1 \cdot r^{n-1}
Write out the formula before plugging anything in.
a6=3261a_6 = 3 \cdot 2^{6-1}
a₁ is 3, r is 2, and n is 6.
a6=325a_6 = 3 \cdot 2^5
Subtract in the exponent first.
a6=332a_6 = 3 \cdot 32
2 to the 5th power is 32.
a6=96a_6 = 96 \checkmark
The 6th term is 96.

The common ratio doesn't have to be a whole number, and it doesn't have to be greater than 1. A ratio between 0 and 1 makes the sequence shrink toward zero. A negative ratio makes the terms alternate between positive and negative.

Consider the sequence 80,20,5,1.25,80, 20, 5, 1.25, \ldots — each term is one-quarter of the previous one, so r = \frac. The sequence is decreasing, but it's still geometric.

Find the 5th term of 80, 20, 5, 1.25, ...
r=2080=14r = \frac{20}{80} = \frac{1}{4}
Divide the second term by the first.
a5=80(14)51a_5 = 80 \cdot \left(\frac{1}{4}\right)^{5-1}
Plug in a₁ = 80, r = 1/4, and n = 5.
a5=80(14)4a_5 = 80 \cdot \left(\frac{1}{4}\right)^4
Simplify the exponent.
a5=801256a_5 = 80 \cdot \frac{1}{256}
(1/4)^4 = 1/256. Raise numerator and denominator separately.
a5=80256=516a_5 = \frac{80}{256} = \frac{5}{16} \checkmark
Simplify the fraction. The sequence is shrinking toward zero.

There's a direct connection between geometric sequences and exponential functions. The explicit formula a_n = a_1 \cdot r^ has exactly the shape of an exponential function: a starting value multiplied by a base raised to a power. If you plotted the terms of a geometric sequence as points on a coordinate plane — with position nn on the x-axis and term value on the y-axis — they would lie exactly on the graph of an exponential function. Geometric sequences are exponential functions whose domain is restricted to positive integers.

Interactive graph — scroll to zoom, drag to pan

The red curve shows the shrinking sequence with r = \frac and a1=80a_1 = 80. The blue curve shows the growing sequence with r=2r = 2 and a1=3a_1 = 3. The actual sequence terms are just the integer inputs on each curve.

Sometimes the problem works backward: you know two terms and need to find rr or write the formula. When you know a1a_1 and one other term, plug both into the explicit formula and solve for rr.

Write the explicit formula for the sequence where a₁ = 5 and a₄ = 135
135=5r41135 = 5 \cdot r^{4-1}
Plug what you know into the explicit formula. You're solving for r.
135=5r3135 = 5 \cdot r^3
Simplify the exponent.
27=r327 = r^3
Divide both sides by 5.
r=273=3r = \sqrt[3]{27} = 3
Take the cube root of both sides to isolate r.
an=53n1a_n = 5 \cdot 3^{n-1} \checkmark
Write the complete formula using a₁ = 5 and r = 3.
Practice Questions
an=a1rn1,a1=2, r=5. Find a4.a_n = a_1 \cdot r^{n-1}, \quad a_1 = 2,\ r = 5.\ \text{Find } a_4.
4, 12, 36, 108,  Find a5.4,\ -12,\ 36,\ -108,\ \ldots\ \text{Find } a_5.
A sequence has a1=6 and a5=96. Find r and write an.\text{A sequence has } a_1 = 6 \text{ and } a_5 = 96.\ \text{Find } r \text{ and write } a_n.
Regents Corner

On Part II and Part III of the Algebra I Regents, geometric sequence problems often ask you to write the explicit formula, find a specific term, or identify whether a sequence is arithmetic or geometric. For full credit, show the value of rr and write the complete formula before substituting — a bare numerical answer without setup earns partial credit at best.

Students write a_n = a_1 \cdot r^n instead of a_n = a_1 \cdot r^, which shifts every answer up by one term. Check your formula with the first term: when n = 1, the formula must return a₁. If you use r^n, you get a₁ · r instead of a₁ — that's the second term, not the first.
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