The arithmetic sequence formula and the minus one inside it
I'm Heron. I coach people through math in their heads, and I once told a box office that row 10 held 48 seats.
My formula had ten jumps in it where the theater has nine. Let's make sure that minus one never costs you anything.
What the arithmetic sequence formula says
An arithmetic sequence adds the same amount at every step. Our rows run 18, 21, 24, 27, and that steady 3 is the common difference, written d.
To reach term n you add d one time fewer than the number suggests. OpenStax writes the formula as aₙ = a₁ + d(n − 1), and everything that goes wrong with it lives in that (n − 1).
Let's walk a row we have not met yet, so the formula does the work for us.
Step 1 of 4the row we want
Row 12
The steps were 11 jumps, 33 seats, and row 12 at 51. Row 1 was full before a single 3 got added.
The one-second check for that minus one
Put n = 1 into your own formula and see what comes back. It has to hand you the first term: 18 + (1 − 1) × 3 = 18, the row we started from.
Write n where (n − 1) belongs and the same test returns 21, a row this theater does not have. One second, and the most common slip in this formula is gone.
The three near misses up top all come from the same gap. 48 adds the 3 ten times where only nine jumps sit between row 1 and row 10, 42 stops at row 9, and 30 counts the jumps while row 1 waits there already full.
Your turn on the same theater.
18, 21, 24 → a₇ =
Same rules as above. You've got this.
The gap has a name: we called it The Fencepost Gap in our math puzzles, where the thing you can count and the thing you were asked for sit one step apart.
Explicit formula or recursive formula?
The explicit formula jumps straight to the term you asked for. Feed aₙ = a₁ + (n − 1)d an n of 50 and row 50 arrives without the 49 rows in front of it.
The recursive formula walks instead. aₙ = aₙ₋₁ + d adds one d to the term before it, which is honest work for row 4 and a long afternoon for row 50.
The sum of an arithmetic sequence
Adding the whole run up gets easier, and the formula says why.Sₙ = n(a₁ + aₙ)/2 is the average of the two ends, times how many terms you have.
Let's book all ten rows and count the seats we just paid for.
Step 1 of 4the two ends
18 and 45
Row 1 and row 10 make 63 together, and so does every pair working inwards. Row 2 with row 9 gives 63, row 3 with row 8 gives 63, and ten rows hold ten of those pairings, which is each row counted twice.
Three questions, and the sequence comes with each one. Count the jumps first, then add.
4, 9, 14 → a₁₀ =
Same rules as above. You've got this.
Now the same sequence, every term added up. The two ends are 4 and the 49 you just found.
4, 9, 14 → S₁₀ =
Same rules as above. You've got this.
One more, and this run falls instead of climbing. A common difference can be negative, and the formula treats it the same.
90, 83, 76 → a₁₂ =
Same rules as above. You've got this.
Where arithmetic sequences show up
Anything that adds the same amount every step is one of these. Rows of seats, a savings plan with a weekly top-up, a running plan that adds two minutes a week.
Count the jumps, not the terms. Row 10 sits nine jumps from row 1, and the formula asks for the jumps.
Test your formula at n = 1. It has to give the first term back, and checking takes a second.
The sum is an average in disguise. Add the ends, halve them, and multiply by how many terms you have.
Honest answer: which part of this formula has burned you?
n − 1
0%
Sum
0%
Both
0%
Neither
0%
FAQ
The formula for the nth term is aₙ = a₁ + (n − 1)d, where a₁ is the first term, d is the common difference, and n is the position you want. OpenStax writes the same thing as a₁ + d(n − 1). The companion formula for the total of the first n terms is Sₙ = n(a₁ + aₙ)/2.
Count the jumps, then add. A theater row that starts at 18 seats and gains 3 every row reaches row 10 after nine jumps, so row 10 holds 18 + 9 × 3 = 45 seats. The jumps are always one fewer than the term number, which is the whole reason the formula subtracts one.
It is the constant amount each step adds, written d. Subtract any term from the one after it and you have it: in 18, 21, 24, 27 the difference is 3 every time. It can be negative, as in 90, 83, 76, where d = −7 and the sequence falls instead of climbing.
The explicit formula, aₙ = a₁ + (n − 1)d, jumps straight to the term you asked for, so term 50 arrives without the 49 terms in front of it. The recursive formula, aₙ = aₙ₋₁ + d, adds one d to the term before it, which is quick for term 4 and slow for term 50.
Average the two ends and multiply by how many terms you have: Sₙ = n(a₁ + aₙ)/2. Ten theater rows running from 18 seats to 45 give (18 + 45) ÷ 2 = 31.5 seats per row on average, and 31.5 × 10 = 315 seats in total.
