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AdvancedGrade 922 min30 / 82

Sequences and arithmetic progressions

General-term and recursive formulas of a sequence; the common difference, nth term and characteristic property of an arithmetic progression, finding it from two terms, and the sum of the first n terms (Gauss’s trick).

Check yourself
In this lesson you will learn
  • Compute the terms of a sequence given by a general-term or a recursive formula
  • Recognise an arithmetic progression and find its difference, any term, and a₁ and d from two terms
  • Use the characteristic property to find missing terms and to check whether numbers form a progression
  • Calculate the sum of the first n terms, including sums of natural and odd numbers

Every week Aysel puts 2 manat more into her money box than the week before: 5 manat in the first week, then 7, 9, 11, … manat. How much will she put in during week 10? In which week will she put in 45 manat? How much will she have saved after 10 weeks? You could count week by week, but for 100 weeks that would take far too long. In this lesson you will learn to answer all such questions with one-line formulas.

Aysel adds the same amount every week, just as the function y = kx + b from the lesson “Linear functions and their graphs” changes in equal steps. An arithmetic progression is that same idea for numbered terms. In the next lesson, “Geometric progressions”, we will multiply by the same number each time instead of adding.

Sequences and ways to define them

Definition
Sequence

A list of numbers in which every natural number n is matched with a number aₙ: a₁, a₂, a₃, …, aₙ, … These numbers are the terms of the sequence, and n is the position (index) of a term. For example, in the sequence of even numbers a₁ = 2, a₂ = 4, a₁₀ = 20. In other words, a sequence is a function whose domain is the natural numbers.

A sequence is usually given in one of two ways. A general-term (explicit) formula expresses aₙ directly through n: you can compute any term, say a₁₀₀, without knowing its neighbours. A recursive formula gives the first term (sometimes the first two) and a rule for getting the next term from the previous ones: then the terms can only be found one after another.

Computing terms

1) Find a₁, a₂, a₃ and a₁₀ for the sequence aₙ = n² − 1.
2) Write the first five terms of the sequence given by the recursive formula a₁ = 2, aₙ₊₁ = 3aₙ − 1.
3) The Fibonacci sequence: a₁ = a₂ = 1, aₙ₊₂ = aₙ₊₁ + aₙ. Write its first eight terms.

Show solution
1) a₁ = 1 − 1 = 0, a₂ = 4 − 1 = 3, a₃ = 9 − 1 = 8, a₁₀ = 100 − 1 = 99. For a₁₀ we did not need the earlier terms.
2) a₂ = 3 · 2 − 1 = 5, a₃ = 3 · 5 − 1 = 14, a₄ = 3 · 14 − 1 = 41, a₅ = 3 · 41 − 1 = 122.
The sequence: 2, 5, 14, 41, 122, …
3) Each term is the sum of the two before it: 1, 1, 2, 3, 5, 8, 13, 21, …
Is this number a term of the sequence?

Are the numbers 125 and 100 terms of the sequence aₙ = 4n + 1?

Show solution
If a number is a term, its position must be a natural number.
4n + 1 = 125 ⇒ 4n = 124 ⇒ n = 31 — a natural number, so 125 = a₃₁.
4n + 1 = 100 ⇒ 4n = 99 ⇒ n = 24.75 — not a natural number, so 100 is not a term.

Arithmetic progressions: the difference and the nth term

Definition
Arithmetic progression

A sequence in which every term from the second on is obtained by adding the same number d to the previous term: aₙ₊₁ = aₙ + d. The number d is the common difference: d = aₙ₊₁ − aₙ (any term minus the one before it). Such a sequence is also called an arithmetic sequence.

The sign of d shows the “direction”: if d > 0 the progression is increasing (3, 7, 11, 15, …; d = 4), if d < 0 it is decreasing (20, 17, 14, 11, …; d = −3), and if d = 0 all terms are equal (5, 5, 5, …). In Aysel’s progression a₁ = 5 and d = 2.

How do we find the nth term without taking every step? Count the steps: a₂ = a₁ + d, a₃ = a₁ + 2d, a₄ = a₁ + 3d. Each time the coefficient of d is one less than the position, because getting from the first term to the nth takes n − 1 steps, and each step adds d.

aₙ = a₁ + (n − 1) · d
where:
  • aₙthe nth term
  • a₁the first term
  • dthe common difference
  • nthe position of the term (a natural number)

The nth term = the first term + (n − 1) steps.

Aysel’s money box

a₁ = 5, d = 2.
1) How much will Aysel put in during week 10?
2) In which week will she put in 45 manat?

Show solution
1) a₁₀ = a₁ + 9d = 5 + 9 · 2 = 23 manat.
2) 5 + (n − 1) · 2 = 45 ⇒ (n − 1) · 2 = 40 ⇒ n − 1 = 20 ⇒ n = 21, so in week 21.
Three more cases

1) Find the 15th term of the progression with a₁ = 12 and d = −4.
2) How many terms does the progression 7, 11, 15, …, 207 have?
3) Which is the first negative term of 50, 46, 42, …?

Show solution
1) a₁₅ = 12 + 14 · (−4) = 12 − 56 = −44.
2) d = 4. 7 + (n − 1) · 4 = 207 ⇒ (n − 1) · 4 = 200 ⇒ n − 1 = 50 ⇒ n = 51 terms.
3) d = −4, aₙ = 50 − 4(n − 1) = 54 − 4n.
54 − 4n < 0 ⇒ n > 13.5 ⇒ n = 14.
a₁₄ = 54 − 56 = −2 (a₁₃ = 2 is still positive).

Often a₁ and d are not given; instead two terms of the progression are known. From aₘ to aₙ there are n − m steps, so we can count from any known term:

aₙ = aₘ + (n − m) · d ⇒ d = (aₙ − aₘ) / (n − m)aₙ = aₘ + (n − m) · d ⇒ d = (aₙ − aₘ) / (n − m)
where:
  • aₘ, aₙtwo known terms (m < n)
  • n − mthe number of steps between them

Difference = difference of the terms ÷ difference of the positions. Then a₁ = aₘ − (m − 1) · d.

Finding a progression from two terms

1) a₃ = 10, a₇ = 22. Find a₁ and d.
2) a₅ = 17, a₁₂ = −4. Find a₁ and d.

Show solution
1) From a₃ to a₇ there are 4 steps: d = (22 − 10) ÷ 4 = 3.
a₁ = a₃ − 2d = 10 − 6 = 4. The progression: 4, 7, 10, 13, 16, 19, 22, …
2) d = (−4 − 17) ÷ (12 − 5) = −21 ÷ 7 = −3.
a₁ = a₅ − 4d = 17 + 12 = 29.
Check: a₁₂ = 29 + 11 · (−3) = −4 ✓

The characteristic property: each term is the mean of its neighbours

Take three consecutive terms aₙ₋₁, aₙ, aₙ₊₁. The middle one is d more than the left one and d less than the right one: aₙ − aₙ₋₁ = aₙ₊₁ − aₙ. Hence 2aₙ = aₙ₋₁ + aₙ₊₁: every term except the first is the arithmetic mean of its neighbours. The converse is also true: if every term of a sequence (from the second on) is the mean of its neighbours, the sequence is an arithmetic progression. That is where the name comes from.

aₙ = (aₙ₋₁ + aₙ₊₁) / 2 aₙ = (aₙ₋ₖ + aₙ₊ₖ) / 2aₙ = (aₙ₋₁ + aₙ₊₁) / 2 aₙ = (aₙ₋ₖ + aₙ₊ₖ) / 2
where:
  • aₙ₋₁, aₙ₊₁the neighbours of aₙ
  • aₙ₋ₖ, aₙ₊ₖthe terms k steps to the left and to the right of aₙ, k < n

A middle term is the mean of the two outer ones — for neighbours and for any terms at equal distance from it.

Using the property

1) In an arithmetic progression a₄ = 11 and a₆ = 19. Find a₅.
2) The numbers x − 1, 2x + 1, 4x − 3 (in this order) form an arithmetic progression. Find x and the numbers.
3) Which of the triples 2, 9, 16 and 3, 8, 14 is an arithmetic progression?

Show solution
1) a₅ = (11 + 19) ÷ 2 = 15.
2) 2(2x + 1) = (x − 1) + (4x − 3) ⇒ 4x + 2 = 5x − 4 ⇒ x = 6.
The numbers: 5, 13, 21 (d = 8).
3) (2 + 16) ÷ 2 = 9 — 2, 9, 16 is an arithmetic progression.
(3 + 14) ÷ 2 = 8.5 ≠ 8 — 3, 8, 14 is not.

The sum of the first n terms: Gauss’s trick

According to a famous story, Carl Friedrich Gauss, who later became a great mathematician, found the sum of the numbers from 1 to 100 very quickly while still a schoolboy. He paired the numbers: 1 + 100, 2 + 99, 3 + 98, … — each pair makes 101, and there are 50 pairs: 50 · 101 = 5050. The same idea works for any arithmetic progression.

Write the sum twice, forwards and backwards: Sₙ = a₁ + a₂ + … + aₙ and Sₙ = aₙ + aₙ₋₁ + … + a₁. Add the terms that stand one above the other: a₁ + aₙ, a₂ + aₙ₋₁, … Moving right, the upper term grows by d and the lower one shrinks by d, so every pair has the same sum, a₁ + aₙ. There are n pairs: 2Sₙ = (a₁ + aₙ) · n.

Interactive
Loading simulation…
Every column is 11 high: 1 + 10 = 2 + 9 = … = 10 + 1. All the columns together make 10 · 11 = 110 — twice the sum, so 1 + 2 + … + 10 = 55.
Sₙ = (a₁ + aₙ) / 2 · nSₙ = (a₁ + aₙ) / 2 · n
where:
  • Sₙthe sum of the first n terms
  • a₁, aₙthe first and the last (nth) term
  • nthe number of terms added

Sum = the mean of the first and last terms × the number of terms.

Sₙ = (2a₁ + (n − 1) · d) / 2 · nSₙ = (2a₁ + (n − 1) · d) / 2 · n
where:
  • a₁, dthe first term and the common difference
  • nthe number of terms

Use it when the last term is unknown: in the first formula replace aₙ by a₁ + (n − 1)d.

Gauss and Aysel

1) Find 1 + 2 + 3 + … + 100 with the formula.
2) How much will Aysel save in 10 weeks (a₁ = 5, d = 2)?

Show solution
1) a₁ = 1, a₁₀₀ = 100, n = 100:
S₁₀₀ = (1 + 100) ÷ 2 · 100 = 50.5 · 100 = 5050.
2) Above we found a₁₀ = 23:
S₁₀ = (5 + 23) ÷ 2 · 10 = 14 · 10 = 140 manat.
Three sum problems

1) Find the sum of the first 20 terms of 7, 4, 1, …
2) Find the sum of all two-digit numbers divisible by 7.
3) The first row of an amphitheatre has 20 seats, and each next row has 2 more seats than the one before. How many seats are there in 15 rows?

Show solution
1) a₁ = 7, d = −3, n = 20:
S₂₀ = (2 · 7 + 19 · (−3)) ÷ 2 · 20 = (14 − 57) · 10 = −430.
2) These numbers form 14, 21, …, 98 (d = 7). Number of terms: (98 − 14) ÷ 7 + 1 = 13.
S₁₃ = (14 + 98) ÷ 2 · 13 = 56 · 13 = 728.
3) a₁ = 20, d = 2: a₁₅ = 20 + 14 · 2 = 48.
S₁₅ = (20 + 48) ÷ 2 · 15 = 34 · 15 = 510 seats.
The reverse problem: how many terms?

How many terms of 3 + 7 + 11 + … must be added to get 210?

Show solution
a₁ = 3, d = 4: Sₙ = (6 + 4(n − 1)) ÷ 2 · n = (2n + 1) · n.
(2n + 1) · n = 210 ⇒ 2n² + n − 210 = 0.
D = 1 + 8 · 210 = 1681 = 41², n = (−1 + 41) ÷ 4 = 10. The other root, −10.5, is not a natural number, so we reject it (see the lesson “Quadratic equations”).
Check: a₁₀ = 39, S₁₀ = (3 + 39) ÷ 2 · 10 = 210 ✓

Sums of natural and odd numbers

Two sums appear so often that it pays to remember them as ready-made formulas. In 1, 2, …, n we have a₁ = 1, aₙ = n; in the odd numbers 1, 3, 5, …, 2n − 1 we have a₁ = 1, aₙ = 2n − 1. Substituting these into the sum formula gives the following.

1 + 2 + 3 + … + n = n(n + 1) / 2 1 + 3 + 5 + … + (2n − 1) = n²1 + 2 + 3 + … + n = n(n + 1) / 2 1 + 3 + 5 + … + (2n − 1) = n²
where:
  • nhow many numbers are added
  • 2n − 1the nth odd number

For even numbers too: 2 + 4 + … + 2n = 2 · n(n + 1)/2 = n(n + 1).

With the ready-made formulas

1) 1 + 2 + … + 50 = ?
2) 1 + 3 + 5 + … + 99 = ?
3) 2 + 4 + 6 + … + 100 = ?

Show solution
1) n = 50: 50 · 51 ÷ 2 = 1275.
2) 2n − 1 = 99 ⇒ n = 50 odd numbers: 50² = 2500.
3) 2 + 4 + … + 100 = 2 · (1 + 2 + … + 50) = 2 · 1275 = 2550.
  1. 1
    Write down what you know

    a₁, d, n, aₙ, Sₙ — which are given and which is asked for? In a word problem, decide what the first term and the step are.

  2. 2
    Choose the formula

    For a term use aₙ = a₁ + (n − 1)d, for a sum one of the Sₙ formulas. If two terms are given, find d first.

  3. 3
    Solve the equation

    Find the unknown. If the unknown is n, the answer must be a natural number; reject fractional or negative roots.

  4. 4
    Check

    Write out the first few terms or recompute the answer with the other formula: the results must agree.

Check yourself: fill in the gap
  1. 1.The common difference of 2, 9, 16, 23, … is d =
  2. 2.If a₁ = 3 and d = 5, then a₁₁ =
  3. 3.If a₆ = 14 and a₈ = 20, then a₇ =
  4. 4.1 + 2 + 3 + … + 20 =
  5. 5.1 + 3 + 5 + … + 19 =
  6. 6.The sum of the first 10 terms of 5, 8, 11, … is S₁₀ =

Key points

  • A sequence is given by a general-term formula aₙ = f(n) or by a recursive formula (the first term + a rule for the next term).
  • Arithmetic progression: aₙ₊₁ = aₙ + d; aₙ = a₁ + (n − 1)d — n − 1 steps from the first term.
  • From two terms: d = (aₙ − aₘ) / (n − m), then a₁ = aₘ − (m − 1)d.
  • Each term is the mean of its neighbours: aₙ = (aₙ₋₁ + aₙ₊₁)/2; if m + n = p + k, then aₘ + aₙ = aₚ + aₖ.
  • Sum: Sₙ = (a₁ + aₙ)/2 · n = (2a₁ + (n − 1)d)/2 · n (Gauss’s trick).
  • 1 + 2 + … + n = n(n + 1)/2; 1 + 3 + … + (2n − 1) = n².

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
What is the common difference of 2, 5, 8, 11, …?