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Arithmetic series

Derive the arithmetic series formula by pairing terms, find unknown sums and term counts, and solve integer threshold problems with original worked practice.

Before you startArithmetic sequences, quadratics and inequalities

01 / A sequence or its sum?

A series adds the terms of a sequence.

The sequence 3, 5, 7, 9 lists individual terms. The series 3 + 5 + 7 + 9 adds them. Write Sₙ for the sum of the first n terms, starting with u₁ = a.

Sₙ = u₁ + u₂ + … + uₙ

Here u₄ = 9, but S₄ = 24. A question about the total needs a sum formula; a question about the amount at one position needs a term formula.

02 / Why the sum formula works

Add the same sum forwards and backwards.

Write the n terms in order, then write another copy underneath in reverse. Every column totals a + l, where l is the last term. There are n columns, and together they add to 2Sₙ.

2Sₙ = n(a + l)
Sₙ = n(a + l)/2

This argument also works for odd n: the central column pairs the middle term with its counterpart in the second copy. It works for n = 1 and for negative or zero differences too.

Since l = a + (n − 1)d, the equivalent form is:

Sₙ = n[2a + (n − 1)d]/2

Pair a sum with its reverseExplore

Each column uses one term from each copy. The column total is always first + last.

S₄ = 4 × 12 ÷ 2 = 24.

The first term is 3. Change the number of terms or the common difference; zero and negative terms are allowed.

Number of terms n 4Common difference d 2

Watch the reversed pairs

Pause, replay or seek freely. The notes explain the same idea and stay in view.

03 / Choose the useful form

Use the information you know.

Add the first 18 terms of 6, 10, 14, …Worked example

a = 6, d = 4, n = 18

The first term is not the common difference.

S₁₈ = 18[12 + 17 × 4]/2 = 720

There are 17 increments between the first and 18th terms.

l = 6 + 17 × 4 = 74

As a check, S₁₈ = 18(6 + 74)/2 = 720.

When first, last and number of terms are already given, n(a + l)/2 is often quickest. If d = 0, every term is a and the formula reduces to Sₙ = na.

04 / Sums of multiples and finite lists

Count the included terms before adding.

Sum the multiples of 7 strictly between 20 and 100Worked example

First = 21, last = 98, d = 7

Both endpoints satisfy the strict bounds.

21 + (n − 1)7 = 98 ⇒ n = 12

The difference of the endpoints counts 11 gaps, not 11 terms.

Sum = 12(21 + 98)/2 = 714

Check the first and last included multiple before using the formula.

If a bound is itself a multiple, “less than” excludes it while “at most” includes it. This changes both the endpoint and the term count.

05 / Find an unknown number of terms

Solve the equation, then apply the index restriction.

How many terms of 5, 8, 11, … total 310?Worked example

Sₙ = n[10 + 3(n − 1)]/2 = n(3n + 7)/2

Substitute a = 5 and d = 3.

3n² + 7n − 620 = 0

Multiply by 2 and rearrange.

n = (−7 ± √7489)/6

Neither value is a positive integer, so no such finite initial sum exists.

A target total does not have to be attainable. Nearby totals here are S₁₃ = 299 and S₁₄ = 343.

If instead the total is 343Worked example

3n² + 7n − 686 = (n − 14)(3n + 49) = 0

Both algebraic roots must be considered.

n = 14 or n = −49/3

Only the positive integer 14 is a valid number of terms.

06 / Recover a and d from sums

Two independent totals give two equations.

S₄ = 38 and S₈ = 124Worked example

S₄ = 2(2a + 3d) = 38 ⇒ 2a + 3d = 19

Divide the first sum equation by 2.

S₈ = 4(2a + 7d) = 124 ⇒ 2a + 7d = 31

Divide the second sum equation by 4.

4d = 12 ⇒ d = 3; a = 5

Subtract to remove a, then substitute back.

Another useful link is uₙ = Sₙ − Sₙ₋₁, with S₀ = 0. For example, if Sₙ = 2n² + 3n, then uₙ = 4n + 1: an arithmetic sequence with a = 5 and d = 4.

07 / Totals crossing a threshold

A decreasing sequence can have sums that rise and then fall.

First sum above 500 for a = 5 and d = 3Worked example

Sₙ = n(3n + 7)/2

Every term is positive, so the partial sums strictly increase.

S₁₇ = 493; S₁₈ = 549

These adjacent values straddle 500.

The least n is 18

The positive-term argument makes this the first crossing.

Why a total can occur twice: a = 12, d = −2Worked example

Sₙ = n(13 − n)

The terms eventually become negative.

Sₙ = 40 ⇒ (n − 5)(n − 8) = 0

Both n = 5 and n = 8 are positive integers, so both are valid.

Sₙ > 40 ⇒ 5 < n < 8

Only n = 6 and n = 7 work. Both sums equal 42.

Partial sums change by Sₙ₊₁ − Sₙ = uₙ₊₁. Positive next terms increase the sum, negative next terms decrease it, and a zero next term leaves it unchanged. Do not discard a second positive integer root automatically.

08 / Patterns and logarithmic sums

Prove the terms are arithmetic before using its sum formula.

The first n odd numbers have a = 1, d = 2, so their sum is n[2 + 2(n − 1)]/2 = n².

Add ln 2 + ln 6 + ln 18 + ln 54 + ln 162Worked example

a = ln 2 and d = ln 3

Each logarithmic difference is ln 3.

S₅ = 5[2ln 2 + 4ln 3]/2 = 5ln 2 + 10ln 3

Use the arithmetic sum formula on the logarithms.

S₅ = ln(2⁵ × 3¹⁰)

For a = 12 and d = −2, find all n with Sₙ ≥ 40. Explain how this differs from Sₙ > 40.

Hint

Factor the quadratic boundary.

Worked solution

Sₙ = n(13 − n), so Sₙ ≥ 40 gives 5 ≤ n ≤ 8: n = 5, 6, 7, 8. A strict inequality excludes n = 5 and 8, leaving 6 and 7.

10 / Recap

Count terms, then choose a sum formula.

Sₙ adds the first n terms; uₙ is just one term.
Use Sₙ = n(a + l)/2 or n[2a + (n − 1)d]/2.
Finite arithmetic lists contain one more term than gaps.
A number of terms must be a positive integer.
Negative terms can make partial sums decrease; retain every valid root.
uₙ = Sₙ − Sₙ₋₁ connects a term rule to a sum rule.

Review arithmetic sequences →

Section 1 of 10 · A sequence or its sum?