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Algebra I/Functions & Exponentials/Arithmetic Sequences
Algebra I Regents in 22 days
Algebra I · Lesson 1

Arithmetic Sequences

How a constant gap between terms connects to linear functions.


Your school cafeteria charges $2.50 per day for lunch. On day 1 you have spent $2.50. On day 2, $5.00. On day 3, $7.50. Each day, the total climbs by exactly $2.50 — the same amount, no exceptions. That fixed, repeating gap is the heart of an arithmetic sequence.

A sequence is a list of numbers in a specific order. Each number in the list is called a term. The first term is a1a_1, the second is a2a_2, and so on. In an arithmetic sequence, you get from one term to the next by adding the same fixed number every single time. That fixed number is called the common difference, and we label it dd.

In the lunch example, the sequence is 2.50,5.00,7.50,10.00,2.50, 5.00, 7.50, 10.00, \ldots and the common difference is d=2.50d = 2.50. To find dd, subtract any term from the term that follows it: 5.002.50=2.505.00 - 2.50 = 2.50. That works for any consecutive pair.

The common difference can be negative. If you start with $40 in a gift card and spend $8 each visit, the balances form the sequence 40,32,24,16,40, 32, 24, 16, \ldots with d=8d = -8. The sequence is decreasing, but the gap is still constant.

Once you know the first term and the common difference, you can find any term without listing every term in between. Think about where each term comes from. The first term is just a1a_1. The second term is a1+da_1 + d. The third is a1+2da_1 + 2d. The fourth is a1+3da_1 + 3d. The pattern: the nnth term has dd added exactly n1n - 1 times. That gives the explicit formula:

an=a1+(n1)da_n = a_1 + (n-1)d

This formula lets you jump straight to the 50th term or the 200th term without computing every term before it. Plug in nn, a1a_1, and dd, and you are done.

Here is a straightforward example. The sequence 3,7,11,15,3, 7, 11, 15, \ldots has a first term of a1=3a_1 = 3 and a common difference of d=4d = 4. Find the 20th term.

Finding the 20th term of 3, 7, 11, 15, ...
an=a1+(n1)da_n = a_1 + (n-1)d
Write the formula first. This keeps your work organized.
a20=3+(201)(4)a_{20} = 3 + (20-1)(4)
Substitute n = 20, a₁ = 3, and d = 4.
a20=3+(19)(4)a_{20} = 3 + (19)(4)
Subtract inside the parentheses first.
a20=3+76a_{20} = 3 + 76
Multiply 19 and 4.
a20=79a_{20} = 79 \checkmark
The 20th term is 79.

Now a harder case. You know two terms of an arithmetic sequence but not the first term or the common difference. The 4th term is 17 and the 9th term is 37. Find the explicit formula.

Finding the formula when two non-consecutive terms are given
a9a4=5da_9 - a_4 = 5d
Going from the 4th term to the 9th term means adding d exactly 5 times — one for each step between them.
3717=5d37 - 17 = 5d
Substitute the values you know.
20=5d20 = 5d
Subtract on the left.
d=4d = 4
Divide both sides by 5.
a1=a43d=173(4)a_1 = a_4 - 3d = 17 - 3(4)
Work backward from the 4th term. You added d three times to get from a₁ to a₄, so subtract it three times to get back.
a1=1712=5a_1 = 17 - 12 = 5
The first term is 5.
an=5+(n1)(4)a_n = 5 + (n-1)(4)
Substitute a₁ = 5 and d = 4 into the explicit formula.
an=5+4n4a_n = 5 + 4n - 4
Distribute the 4.
an=4n+1a_n = 4n + 1 \checkmark
Combine constants. This is the simplified explicit formula.

Look at that final formula: an=4n+1a_n = 4n + 1. Compare it to the slope-intercept form of a line: y=mx+by = mx + b. The common difference d=4d = 4 plays the same role as slope — it is the constant rate of change. The expression a1+(n1)da_1 + (n-1)d, when distributed and simplified, always produces a linear expression in nn. Arithmetic sequences and linear functions are the same mathematical structure. A sequence is just a linear function whose input is restricted to positive whole numbers.

The graph below plots the terms of an=4n+1a_n = 4n + 1 as discrete points, alongside the line y=4x+1y = 4x + 1. The points sit exactly on the line, but the sequence only lives at integer values of nn.

Interactive graph — scroll to zoom, drag to pan
Practice Questions
an=? for 6,10,14,18,, n=12a_n = ?\ \text{for}\ 6, 10, 14, 18, \ldots,\ n = 12
an=? for a1=20, d=3, n=8a_n = ?\ \text{for}\ a_1 = 20,\ d = -3,\ n = 8
a5=3, a10=23. Find a1 and an.a_5 = 3,\ a_{10} = 23.\ \text{Find}\ a_1\ \text{and}\ a_n.
Regents Corner

On Part II and Part III of the Algebra I Regents, arithmetic sequence problems often ask you to write the explicit formula and then evaluate it for a specific term. Both steps are required for full credit. Writing only the formula or only the answer earns partial credit at best. Show the substitution step clearly — the grader needs to see that you used the formula correctly, not just that you produced the right number.

Students substitute n instead of n - 1 and write a₂₀ = a₁ + 20d instead of a₁ + 19d. The 20th term uses d exactly 19 times because you start at a₁ and take 19 steps to reach a₂₀. A reliable check: plug in n = 1 and confirm you get a₁. If a₁ + (1)d comes out instead of a₁, the formula is wrong.
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