Pair the ends
A series is the sum of the terms of a sequence. The terms of an arithmetic sequence go up by the same amount each time, and there is a quick way to add them.
Take 1 + 2 + 3 + … + 10. Pair the first term with the last: 1 + 10 = 11. Then pair the second with the second-last: 2 + 9 = 11. Then 3 + 8 = 11. Every pair makes the same 11, because moving one place inward adds 1 to the left number and takes 1 from the right number, so the total of the pair does not change.
Count the pairs
Ten terms make five pairs: 1 + 10, 2 + 9, 3 + 8, 4 + 7 and 5 + 6. Each pair is worth 11, so the sum is 5 × 11 = 55.
Here is the same idea drawn. Write the sum twice, once forward and once backward, and put the two copies side by side. Each of the 10 places holds 11, so the two copies add to 10 × 11 = 110, and one copy is half of that: 55.
1 + 2 + … + n is a staircase; a second copy turned upside down fills its gaps
Set n = 6 and slide the second staircase in
The gold staircase is 1 + 2 + … + 6. Slide in a second copy, turned upside down: the two fill a 6 by 7 rectangle of 42 blocks, so one staircase holds 42 ÷ 2 = 21.
The formula
Any arithmetic series pairs up the same way. As one end rises by the common difference, the other end falls by the same amount, so every pair is worth the first term plus the last. With n terms there are pairs, so the sum is .
If n is odd, the term in the middle has no partner. It lies halfway between the first and last terms, so it is worth exactly half a pair, and the formula still works: .
With first term a and last term l, the sum of n terms is . When the last term is not given, use l = a + (n − 1)d, where d is the common difference, as in The nth Term. Then the sum is .
Ten terms from 2 to 29
Find 2 + 5 + 8 + … + 29. First count the terms. They go up by 3, and from 2 to 29 is 27, which is 27 ÷ 3 = 9 steps of 3. Nine steps join ten terms, so there are 10 terms.
Ten terms make 10 ÷ 2 = 5 pairs, and each pair is worth 2 + 29 = 31. So the sum is 5 × 31 = 155.
The five pairs of 2 + 5 + 8 + … + 29. Each is worth 31, so the sum is 5 × 31 = 155.
Page numbers
In the next problem, a magazine has pages numbered 1 to 40. Their sum comes from the same pairing: 40 pages make 20 pairs, each worth 1 + 40 = 41.
Worked example: Torn-Out Sheet / Missing Page Numbers
Question A student accidentally tears out 1 single sheet of paper from a magazine that originally contained 40 consecutive numbered pages (pages 1 to 40). The sum of all the page numbers on the remaining pages is 749. What are the two page numbers printed on the torn-out sheet?
1.Sum of 40 pages using pairing: 20 pairs of (1 + 40 = 41) ⟹ 20 × 41 = 820.
Pair the pages from both ends: 1 + 40, 2 + 39, ... twenty pairs of 41, so all 40 pages sum to 820. 2.Missing page sum = 820 − 749 = 71.
The remaining pages sum to 749, so the torn sheet carried 820 − 749 = 71. 3.Draw two comparison bars for the two sides of the single sheet: Front page [u], Back page [u][1].
A sheet has two consecutive page numbers: the front is one unit, the back one unit and 1 more. 4.Value of 2u = 71 − 1 = 70 ⟹ u = 35.
2u + 1 = 71, so 2u = 70 and u = 35. 5.Front page = 35, Back page = 35 + 1 = 36.
The torn sheet is pages 35 and 36.
Answer: Pages 35 and 36
Common mistakes
- Assuming a torn sheet can carry an even page on the front and an odd page on the back (e.g., pages 34 and 35), violating standard book pagination where recto is odd and verso is even.
- Using 40 × 40 = 1600 instead of the triangular summation formula n(n+1)2.
- Dividing 71 by 2 and rounding arbitrarily without checking consecutive integers.
More patterns and page numbers problems, worked step by step →
A series in sigma notation
A sum written with is often an arithmetic series. The sum has the terms 24, 26, 28, …, 82, which go up by 2. It can be added with . It can also be split into two simpler sums: 22 added once for each of the 30 values of r, and 2 times . By the pairing, is .
Worked example: The Seats in a Stadium Stand Written in Sigma Notation, and the Seats Behind the Reserved Rows
Question A stand at a stadium has 30 rows of seats. Row r, counted from the front, has (22 + 2r) seats, so the front row has 24 seats. (a) Write the total number of seats in the stand in sigma notation, and evaluate it. (b) Rows 1 to 10 are reserved for season-ticket holders. Write the number of seats in rows 11 to 30 in sigma notation, and evaluate it.
1.Row r has (22 + 2r) seats, and r runs from 1 to 30. So the total is ∑r=130 (22 + 2r).
The total is ∑r=130 (22 + 2r): one term for each row. 2.Split the sum into ∑r=130 22 + 2∑r=130 r. The first part adds 22 once for each of the 30 rows, which is 30 × 22 = 660. The second part is 2 × 30 × 312 = 930.
Split it: ∑r=130 22 = 660 and 2∑r=130 r = 2 × 465 = 930. 3.(a) The stand has 660 + 930 = 1590 seats. Check with the first and last rows: row 30 has 22 + 60 = 82 seats, and 302(24 + 82) = 15 × 106 = 1590.
(a) The stand has 1590 seats. 4.The seats behind the reserved rows are ∑r=1130 (22 + 2r). This is the whole sum less rows 1 to 10, and ∑r=110 (22 + 2r) = 10 × 22 + 2 × 10 × 112 = 220 + 110 = 330.
The gold rows are ∑r=1130 (22 + 2r): the whole sum less ∑r=110 (22 + 2r) = 330. 5.(b) ∑r=1130 (22 + 2r) = 1590 − 330 = 1260 seats. Check: rows 11 to 30 are 20 rows, from 44 seats up to 82 seats, and 202(44 + 82) = 10 × 126 = 1260.
(b) Rows 11 to 30 have 1590 − 330 = 1260 seats.
Answer: (a) ∑r=130 (22 + 2r) = 1590 seats; (b) ∑r=1130 (22 + 2r) = 1260 seats
Common mistakes
- Evaluating ∑r=130 22 as 22. The 22 is added once for every value of r, so it contributes 30 × 22 = 660.
- Taking away the first 11 rows instead of the first 10. The sum from row 11 to row 30 includes row 11, so the rows taken away are 1 to 10. Counting the rows that are left, 30 − 10 = 20, catches the slip.