Algebra 2

Arithmetic series and partial sums

The sequence 13, 19, 25, and onward climbs by 6 each time. What do its first 25 terms add up to, without doing 24 additions?

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Sigma notationAlgebra 2

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Arithmetic series and partial sums

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Why this exists

The sequence 13, 19, 25, 31, \ldots climbs by 6 at every step. Its terms are easy to make, and its 25th term is easy to find with the explicit formula: 13 + 24 \cdot 6 = 157. But a sequence is often wanted for its total: every seat in a stand, every can in a display, all the money saved over half a year. Adding 25 terms one at a time takes 24 additions, each one a chance to slip, and 250 terms would take 249.

Sigma notation writes such a total exactly,

13 + 19 + 25 + \cdots + 157 = \sum_{k=1}^{25} (6k + 7)

but writing a sum down is not the same as finding it. This lesson finds it. For an arithmetic sequence the whole total comes from three numbers, the first term, the last term and how many terms there are, in one multiplication.

Definition

The nth partial sum of a sequence a_1, a_2, a_3, \ldots is the total of its first n terms. It is written S_n:

S_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^{n} a_k

“Partial” because it stops after n terms; the sequence itself may go on forever. The partial sums form a sequence of their own, the running totals. For 13, 19, 25, \ldots they are

\begin{aligned} S_1 &= 13 \\ S_2 &= 13 + 19 = 32 \\ S_3 &= 32 + 25 = 57 \end{aligned}

and so on, each running total being the one before plus the next term.

An arithmetic series is the sum of terms of an arithmetic sequence, one in which each term is the one before plus the same common difference d. Its partial sums are what this lesson is about. “Series” is the word for a sum of the terms of a sequence; the sequence is the list, the series is the list added up.

Intuition

Write the sum twice, once forwards and once backwards, and add the two lists column by column. Figure 1 does this for 5 + 11 + 17 + 23 + 29.

Figure 1The terms 5, 11, 17, 23, 29 (shaded) and the same terms in reverse (dashed). Every column is 5 + 29 = 34, so two copies of the sum fill 5 \times 34 = 170, and one copy is 85.

In the first column, 5 meets 29. Move one column right and the forwards term goes up by 6 while the backwards term goes down by 6, so their total does not change: 11 + 23 = 34, then 17 + 17 = 34, and so on. Every column is 34, the first term plus the last.

So two copies of the sum make a rectangle, 5 columns wide and 34 high, of area 170. One copy is half of that:

5 + 11 + 17 + 23 + 29 = \frac{5 \times 34}{2} = 85

The same picture says something else. Half of 34 is 17, the middle term, so the sum is 5 \times 17: five terms, each “on average” 17. The terms below the middle are short of it by exactly as much as the terms above it are over, so an arithmetic series adds up as if every term were the average of the first and the last.

Results

The sum of an arithmetic series. For an arithmetic sequence with first term a_1 and nth term a_n,

S_n = \frac{n\,(a_1 + a_n)}{2}

In words: the number of terms times the average of the first and last terms.

The same sum from a_1 and d. Putting a_n = a_1 + (n - 1)d into the first form gives

S_n = \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr)

Use the first form when the last term is known, the second when the common difference is.

The sum of the first n counting numbers. With a_1 = 1 and a_n = n,

1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}

So \displaystyle 1 + 2 + \cdots + 100 = \frac{100 \cdot 101}{2} = 5050. In sigma notation this is \displaystyle \sum_{k=1}^{n} k = \frac{n(n + 1)}{2}, the short formula the properties of sums were waiting for.

The sum of the first n odd numbers. With a_1 = 1 and a_n = 2n - 1,

1 + 3 + 5 + \cdots + (2n - 1) = \frac{n \cdot 2n}{2} = n^2

The partial sums are quadratic in n. Multiplying out the second form,

S_n = \frac{d}{2}\,n^2 + \left(a_1 - \frac{d}{2}\right)n

a quadratic in n with no constant term (when d \ne 0). For 13, 19, 25, \ldots this is S_n = 3n^2 + 10n.

Terms from partial sums. Each running total is the one before plus one more term, so

a_1 = S_1, \qquad a_n = S_n - S_{n-1} \quad \text{for } n \ge 2

This holds for every sequence, arithmetic or not.

Why it is true

Write the sum forwards and then backwards. Both lines are the same n numbers, so both equal S_n:

\begin{aligned} S_n &= a_1 + a_2 + \cdots + a_n \\ S_n &= a_n + a_{n-1} + \cdots + a_1 \end{aligned}

Add the two lines column by column. The commutative and associative properties of addition allow it: all 2n numbers can be regrouped into n pairs, one from each line.

Take the kth column. The term in the top line is k - 1 steps after the first term, a_k = a_1 + (k - 1)d. The term in the bottom line is k - 1 steps before the last term, a_{n+1-k} = a_n - (k - 1)d. Their sum is

\begin{aligned} &\bigl(a_1 + (k - 1)d\bigr) + \bigl(a_n - (k - 1)d\bigr) \\ &\quad = a_1 + a_n \end{aligned}

The (k - 1)d cancels, so every column has the same total, a_1 + a_n. There are n columns, so

2S_n = n\,(a_1 + a_n)

and dividing by 2 gives \displaystyle S_n = \frac{n(a_1 + a_n)}{2}. The cancellation is the whole proof, and it needs the steps to be equal: one line climbs by d exactly as fast as the other falls by d.

For the second form, substitute the explicit formula for the last term:

\begin{aligned} S_n &= \frac{n}{2}\bigl(a_1 + a_1 + (n - 1)d\bigr) \\ &= \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr) \end{aligned}

The rule a_n = S_n - S_{n-1} needs no algebra: S_n is a_1 + \cdots + a_{n-1}, which is S_{n-1}, plus one more term, a_n.

Worked examples

Example 1

The first 25 terms

Find the sum of the first 25 terms of 13, 19, 25, \ldots

Solution

The terms rise by 6, so the sequence is arithmetic with a_1 = 13 and d = 6. The first form needs the last term, the 25th, which is 24 steps after the first:

a_{25} = 13 + 24 \cdot 6 = 157

Now there are 25 terms, the first is 13 and the last is 157:

S_{25} = \frac{25\,(13 + 157)}{2} = \frac{25 \cdot 170}{2} = 25 \cdot 85 = 2125

The second form gives the same total without finding a_{25} first:

S_{25} = \frac{25}{2}\bigl(2 \cdot 13 + 24 \cdot 6\bigr) = \frac{25}{2} \cdot 170 = 2125

The average term is 85, halfway between 13 and 157, and 25 terms of average 85 total 2125.

Example 2

A sum in sigma notation

Find \sum_{k=1}^{40} (5k - 2).

Solution

The rule 5k - 2 is linear in k, so each term is 5 more than the one before: the terms form an arithmetic sequence with d = 5. Putting in the limits gives the first and last terms:

k = 1: \ 5 - 2 = 3 \qquad k = 40: \ 200 - 2 = 198

There are 40 - 1 + 1 = 40 terms, so

\sum_{k=1}^{40} (5k - 2) = \frac{40\,(3 + 198)}{2} = 20 \cdot 201 = 4020

Check with the properties of sums and the formula for \sum k. The sum splits into 5\sum_{k=1}^{40} k - \sum_{k=1}^{40} 2, and \displaystyle \sum_{k=1}^{40} k = \frac{40 \cdot 41}{2} = 820:

5 \cdot 820 - 2 \cdot 40 = 4100 - 80 = 4020

Example 3

Counting the terms first

Find 47 + 43 + 39 + \cdots + (-25).

Solution

The terms fall by 4, so the sum is arithmetic with a_1 = 47 and d = -4. The last term is given, but the number of terms is not, and the formula cannot be used without it. The term in position n is

a_n = 47 - 4(n - 1) = 51 - 4n

and the last term is in the position where 51 - 4n = -25, so 4n = 76 and n = 19. Then

S_{19} = \frac{19\,\bigl(47 + (-25)\bigr)}{2} = \frac{19 \cdot 22}{2} = 209

The negative terms are allowed for. Splitting the sum shows it: the twelve positive terms run from 47 down to 3 and total \displaystyle \frac{12 \cdot 50}{2} = 300, and the seven negative terms run from -1 to -25 and total \displaystyle \frac{7 \cdot (-26)}{2} = -91. And 300 - 91 = 209.

Example 4

When the total is the target

Oksana saves 20 dollars in the first week, and in each later week she saves 3 dollars more than the week before. (a) How much has she saved in total after 20 weeks? (b) In which week does her total first reach 1000 dollars?

Solution

The weekly amounts 20, 23, 26, \ldots are arithmetic with a_1 = 20 and d = 3, and her total after n weeks is the partial sum S_n.

(a) In week 20 she saves a_{20} = 20 + 19 \cdot 3 = 77 dollars, so

S_{20} = \frac{20\,(20 + 77)}{2} = 10 \cdot 97 = 970

After 20 weeks she has 970 dollars.

(b) Now the total is known and n is not. The second form gives S_n in terms of n alone:

S_n = \frac{n}{2}\bigl(40 + 3(n - 1)\bigr) = \frac{n(3n + 37)}{2}

Setting this equal to 1000 gives 3n^2 + 37n - 2000 = 0. By the quadratic formula

\begin{aligned} n &= \frac{-37 + \sqrt{37^2 + 24000}}{6} \\ &= \frac{-37 + \sqrt{25369}}{6} \approx 20.4 \end{aligned}

taking the positive root, since a number of weeks is positive. This is not a whole number, so no week ends with exactly 1000 dollars. The total is 970 after week 20 and, with a_{21} = 80, it is \displaystyle \frac{21(20 + 80)}{2} = 1050 after week 21. Her total first reaches 1000 dollars in week 21.

Example 5

From the running totals back to the terms

The partial sums of a sequence are S_n = 3n^2 + 10n. Find the terms, and show that the sequence is arithmetic.

Solution

The first term is the first running total: a_1 = S_1 = 3 + 10 = 13. For n \ge 2, each term is what the running total gains at step n:

\begin{aligned} a_n &= S_n - S_{n-1} \\ &= 3n^2 + 10n - \bigl(3(n - 1)^2 + 10(n - 1)\bigr) \\ &= 3n^2 + 10n - 3n^2 + 6n - 3 - 10n + 10 \\ &= 6n + 7 \end{aligned}

At n = 1 the rule 6n + 7 gives 13, which is a_1, so it holds for every term. A rule of the form 6n + 7 rises by exactly 6 whenever n rises by 1, so the sequence is arithmetic with d = 6: it is 13, 19, 25, \ldots, the sequence of Example 1. As a check, S_{25} = 3 \cdot 625 + 10 \cdot 25 = 1875 + 250 = 2125, as found there.

Example 6

Multiples of 7

Find the sum of all the three-digit multiples of 7.

Solution

The multiples of 7 rise by 7, so they form an arithmetic sequence. The task is to find where the three-digit ones start and stop. Since 7 \cdot 14 = 98 has two digits and 7 \cdot 15 = 105 has three, the first is 105. Since 7 \cdot 142 = 994 and 7 \cdot 143 = 1001 has four digits, the last is 994.

The multiples run from 7 \cdot 15 to 7 \cdot 142, one for each whole number from 15 to 142, so there are 142 - 15 + 1 = 128 of them. Then

S = \frac{128\,(105 + 994)}{2} = 64 \cdot 1099 = 70336

The same answer comes from factoring out the 7: the sum is 7(15 + 16 + \cdots + 142), and \displaystyle 15 + \cdots + 142 = \frac{128 \cdot 157}{2} = 10048, with 7 \cdot 10048 = 70336.

Example 7

Two partial sums, two unknowns

The first 10 terms of an arithmetic sequence add to 155, and the first 20 add to 610. Find the first term and the common difference.

Solution

Each total gives an equation through the second form of the formula. With n = 10,

\frac{10}{2}(2a_1 + 9d) = 155

so 2a_1 + 9d = 31, and with n = 20,

\frac{20}{2}(2a_1 + 19d) = 610

so 2a_1 + 19d = 61.

Subtracting the first equation from the second removes a_1: 10d = 30, so d = 3. Then 2a_1 + 27 = 31 gives a_1 = 2. The sequence is 2, 5, 8, \ldots

Check. a_{10} = 2 + 27 = 29 and \displaystyle S_{10} = \frac{10(2 + 29)}{2} = 155; a_{20} = 2 + 57 = 59 and \displaystyle S_{20} = \frac{20(2 + 59)}{2} = 610.

Boundaries

The formula needs a common difference. The pairing works only because one line climbs by d exactly as the other falls by d. For 1 + 2 + 4 + 8 the pairs are 1 + 8 = 9 and 2 + 4 = 6, and \displaystyle \frac{4(1 + 8)}{2} = 18 is wrong; the sum is 15. Sums of geometric sequences have their own formula, in Geometric series.

n is the number of terms, not the last term or the last index. From position m to position p there are p - m + 1 terms. For a listed sum, n comes from solving a_1 + (n - 1)d = a_n, and it must come out a positive whole number; if it does not, the last number given is not a term of the sequence.

A target total may fall between two partial sums. Solving S_n = T for n gives a quadratic, and its root may not be a whole number. Then no partial sum equals T, and the question becomes which partial sum first passes it, as in Example 4. A negative root is never a number of terms.

The sum is finite. The formula adds the first n terms. An arithmetic series that goes on forever has no total unless every term is 0: if d \ne 0 the terms eventually all have one sign and grow without limit in size, and if d = 0 with a_1 \ne 0 every term is the same non-zero number. Either way the partial sums grow without limit in size. Totals of infinitely many terms are taken up in Infinite series and partial sums.

A common difference of 0 is allowed. Then every term is a_1, the formula gives \displaystyle \frac{n(a_1 + a_1)}{2} = n a_1, and that is right. The partial sums are then linear in n rather than quadratic.

Where this goes next

The counting-number sum 1 + 2 + \cdots + n has a picture of its own: a triangle of dots, row k holding k dots, gives the triangular numbers 1, 3, 6, 10, \ldots, and two such triangles fit together into a rectangle exactly as in Figure 1. Formulas like \displaystyle \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} are also the first place where proof by induction earns its keep, and Induction on a sum formula proves sum formulas, including ones for \sum k^2 that no pairing trick reaches, by adding one more term to a case already known.

Changing “add the same number” to “multiply by the same number” gives geometric series, the sums behind loan repayments and repeated deposits, which need a different trick. And asking what happens to S_n as n grows without limit is the start of infinite series, where a sum of infinitely many terms is defined as the limit of its partial sums. An arithmetic series never has such a limit unless it is all zeros; a geometric one, as Infinite geometric series shows, sometimes does.

Exercises

Sums from the formula

Find the sum of the first 30 terms of 6, 15, 24, \ldots
Find the sum of the first 16 terms of 50, 47, 44, \ldots
Find the sum of the first 20 terms of \displaystyle \frac{1}{2}, \frac{5}{4}, 2, \ldots
Find 1 + 2 + 3 + \cdots + 200.
Find the sum of the first 50 odd numbers.

Sums in sigma notation

Find \sum_{k=1}^{60} (3k + 4).
Find \sum_{k=1}^{24} (40 - 3k).
Find \sum_{k=5}^{30} (4k - 1).

Counting the terms first

Find 8 + 14 + 20 + \cdots + 206.
Find (-15) + (-11) + (-7) + \cdots + 61.
Find the sum of all the multiples of 6 between 100 and 400.

In a setting

A display of cans has 30 cans in its bottom row, and each row above has 2 fewer than the row below it. The top row has 6 cans. How many cans are in the display?
A clock strikes once at one o’clock, twice at two o’clock, and so on up to twelve times at twelve o’clock, and strikes only on the hour. How many times does it strike in a full day of 24 hours?
A runner runs 2 km on the first day of a plan and 0.5 km more on each day after. How far does she run in total over the 28 days of the plan?
How many terms of 7 + 10 + 13 + \cdots add up to 282?

Working backwards

An arithmetic series has 18 terms, first term 7 and sum 891. Find its last term and its common difference.
The partial sums of a sequence are S_n = 2n^2 - 5n. Find a_1 and a formula for a_n, and show that the sequence is arithmetic.
For an arithmetic sequence, S_8 = 100 and S_{16} = 328. Find a_1 and d.

Reasoning

Show that pairing the first and last terms does not work for 1 + 3 + 9 + 27, and explain what property of an arithmetic sequence the pairing relies on.
Use the sum formula to prove that the first n odd numbers add to n^2.
Show that when n is odd, the sum of the first n terms of an arithmetic sequence is n times the middle term.
A classmate says the first 2n terms of an arithmetic sequence add to twice the first n. Show with 13, 19, 25, \ldots that this is false, and find S_{2n} - 2S_n in general. When is the classmate right?
Give a second proof of \displaystyle S_n = \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr) using the properties of sums: write S_n = \sum_{k=1}^{n} \bigl(a_1 + (k - 1)d\bigr), split it, and use \displaystyle \sum_{k=1}^{n} (k - 1) = \frac{(n - 1)n}{2}. Say where that last formula comes from.

Answers

4095. Here a_1 = 6 and d = 9, so a_{30} = 6 + 29 \cdot 9 = 267 and \displaystyle S_{30} = \frac{30(6 + 267)}{2} = 15 \cdot 273 = 4095.
440. d = -3, so a_{16} = 50 - 15 \cdot 3 = 5 and \displaystyle S_{16} = \frac{16(50 + 5)}{2} = 440.
\displaystyle \frac{305}{2}. \displaystyle d = \frac{3}{4}, so \displaystyle a_{20} = \frac{1}{2} + 19 \cdot \frac{3}{4} = \frac{59}{4} and \displaystyle S_{20} = 10\left(\frac{2}{4} + \frac{59}{4}\right), which is \displaystyle \frac{610}{4} = \frac{305}{2}.
20100. \displaystyle \frac{200 \cdot 201}{2} = 20100.
2500. The first n odd numbers add to n^2, and 50^2 = 2500; directly, the last is 99 and \displaystyle \frac{50(1 + 99)}{2} = 2500.
5730. The terms run from 7 to 184, so the sum is \displaystyle \frac{60(7 + 184)}{2} = 30 \cdot 191 = 5730.
60. The terms run from 37 down to 40 - 72 = -32, so the sum is \displaystyle \frac{24(37 - 32)}{2} = 60.
1794. There are 30 - 5 + 1 = 26 terms, from 19 to 119, so the sum is \displaystyle \frac{26(19 + 119)}{2} = 13 \cdot 138 = 1794.
3638. d = 6 and 8 + 6(n - 1) = 206 gives n = 34; then \displaystyle \frac{34(8 + 206)}{2} = 17 \cdot 214 = 3638.
460. d = 4 and -15 + 4(n - 1) = 61 gives n = 20; then \displaystyle \frac{20(-15 + 61)}{2} = 460.
12450. The multiples run from 102 = 6 \cdot 17 to 396 = 6 \cdot 66, so there are 66 - 17 + 1 = 50 of them, and \displaystyle \frac{50(102 + 396)}{2} = 25 \cdot 498 = 12450.
234 cans. The rows 30, 28, \ldots, 6 fall by 2, so there are \displaystyle \frac{30 - 6}{2} + 1 = 13 rows, and \displaystyle \frac{13(30 + 6)}{2} = 234.
156 times. Each twelve hours it strikes \displaystyle 1 + 2 + \cdots + 12 = \frac{12 \cdot 13}{2} = 78 times, and a day has two such spans: 2 \cdot 78 = 156.
245 km. On day 28 she runs 2 + 27 \cdot 0.5 = 15.5 km, so the total is \displaystyle \frac{28(2 + 15.5)}{2} = 14 \cdot 17.5 = 245.
12 terms. \displaystyle S_n = \frac{n}{2}\bigl(14 + 3(n - 1)\bigr) = \frac{n(3n + 11)}{2}, and setting it to 282 gives 3n^2 + 11n - 564 = 0, so \displaystyle n = \frac{-11 + 83}{6} = 12 (the other root is negative). Check: a_{12} = 40 and \displaystyle \frac{12(7 + 40)}{2} = 282.
Last term 92, common difference 5. \displaystyle \frac{18(7 + a_{18})}{2} = 891 gives 7 + a_{18} = 99, so a_{18} = 92; the 17 steps from 7 to 92 climb 85, so d = 5.
a_1 = -3 and a_n = 4n - 7, with d = 4. a_1 = S_1 = 2 - 5 = -3, and for n \ge 2, S_n - S_{n-1} is 2n^2 - 5n - 2(n - 1)^2 + 5(n - 1), which simplifies to 4n - 7. That rule also gives -3 at n = 1. Each term is 4 more than the one before, so the sequence is arithmetic.
\displaystyle a_1 = \frac{11}{2} and d = 2. 4(2a_1 + 7d) = 100 gives 2a_1 + 7d = 25, and 8(2a_1 + 15d) = 328 gives 2a_1 + 15d = 41. Subtracting, 8d = 16, so d = 2 and 2a_1 = 11.
The pairs are 1 + 27 = 28 and 3 + 9 = 12, which differ, so the sum is not “number of pairs times one pair”: \displaystyle \frac{4(1 + 27)}{2} = 56, but the true sum is 40. The pairing relies on a common difference. Moving one column inwards adds d to the front term and takes d from the back term, so the pair total stays the same. Here the steps 2, 6, 18 are unequal.
The odd numbers 1, 3, 5, \ldots are arithmetic with a_1 = 1 and d = 2, so the nth is 1 + 2(n - 1) = 2n - 1. Then \displaystyle S_n = \frac{n\bigl(1 + (2n - 1)\bigr)}{2}, which is \displaystyle \frac{n \cdot 2n}{2} = n^2.
When n is odd the middle term is in position \displaystyle m = \frac{n + 1}{2}, which is \displaystyle m - 1 = \frac{n - 1}{2} steps from the start. So \displaystyle a_m = a_1 + \frac{n - 1}{2}d, which is \displaystyle \frac{2a_1 + (n - 1)d}{2}, and that is \displaystyle \frac{a_1 + a_n}{2}. Then \displaystyle S_n = \frac{n(a_1 + a_n)}{2} = n\,a_m.
With S_n = 3n^2 + 10n, S_1 = 13 but S_2 = 32, not 26. In general \displaystyle S_n = \frac{d}{2}n^2 + \left(a_1 - \frac{d}{2}\right)n. Doubling n makes the first part 2dn^2 and doubling S_n makes it dn^2, while the linear parts are equal and cancel, so S_{2n} - 2S_n = dn^2. The classmate is right only when d = 0. With d > 0 the second n terms are larger than the first n, so they add to more.
Split the sum and take out the constants. The sum becomes \sum_{k=1}^{n} a_1 + d\sum_{k=1}^{n} (k - 1), which is \displaystyle n a_1 + d \cdot \frac{(n - 1)n}{2}, and that is \displaystyle \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr). The last formula is the counting-number sum: \sum_{k=1}^{n} (k - 1) = 0 + 1 + \cdots + (n - 1), which is \displaystyle \frac{(n - 1)n}{2} by the formula for 1 + 2 + \cdots + m with m = n - 1.