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Working backwards

Patterns and sequences · level 5, lesson 2 of 6

A question to start

Someone wrote out an arithmetic sequence and then spilled water on the page. Only two numbers survive:

the 5th term is 17, and the 12th term is 45.

Can you rebuild the whole sequence, every term, from just these two? And here is a sharper question: is 300 somewhere in that sequence? You will not find it by writing out terms for a long time. One short calculation answers it for sure.

The idea

Two numbers decide everything

In lesson 1 you saw that an arithmetic sequence is fixed completely by two numbers: the first term \(a\) and the common difference \(d\). So if we can find \(a\) and \(d\), we know every term. The surprise is that any two terms, not only the first two, are enough to find them.

Step 1: count the jumps between the two terms

From the 5th term to the 12th term there are \(12 - 5 = 7\) jumps of size \(d\).

t₅ = 17 t₁₂ = 45 7 jumps of d 7d = 45 − 17 = 28

The terms climbed by \(45 - 17 = 28\) in those 7 jumps, so each jump is

\[d = \frac{45 - 17}{12 - 5} = \frac{28}{7} = 4.\]

In general, for any two terms \(t_m\) and \(t_k\) (with \(m < k\)):

\[d = \frac{t_k - t_m}{k - m} = \frac{\text{change in term}}{\text{change in position}}\]

If you have met the slope of a line, this is exactly the slope between the two points \((m, t_m)\) and \((k, t_k)\). That makes sense: the points of an arithmetic sequence lie on a straight line whose slope is \(d\).

Step 2: walk back to the first term

The 5th term is 4 jumps after the first, so going back 4 jumps from 17:

\[a = 17 - 4 \times 4 = 1.\]

The sequence is 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, ... and the general term is

\[t_n = 1 + (n - 1) \times 4 = 4n - 3.\]

Check both given terms: \(t_5 = 20 - 3 = 17\) ✓ and \(t_{12} = 48 - 3 = 45\) ✓.

The same thing with algebra

You can also write each given fact as an equation using \(t_n = a + (n - 1)d\):

\[a + 4d = 17 \qquad\qquad a + 11d = 45\]

Subtract the first equation from the second. The \(a\)'s cancel, leaving \(7d = 28\), so \(d = 4\), and then \(a = 17 - 16 = 1\). This is a system of two equations: two unknowns need two facts. It is the same calculation as the picture, written in symbols. The picture explains why subtracting works: subtracting removes the shared start \(a\) and leaves only the jumps between the two terms.

Is a number in the sequence?

Is 300 a term of \(t_n = 4n - 3\)? If it is, it sits at some whole-number position \(n\). Solve:

\[4n - 3 = 300 \quad\Rightarrow\quad 4n = 303 \quad\Rightarrow\quad n = 75.75.\]

There is no 75.75th term, so 300 is not in the sequence. It falls between \(t_{75} = 297\) and \(t_{76} = 301\).

The general test: set \(t_n\) equal to the number and solve for \(n\). If \(n\) is a whole number of at least 1, the number is a term (and \(n\) tells you which one). If \(n\) is a fraction, zero or negative, it is not.

There is a neat way to see the answer without solving: every term of 1, 5, 9, 13, ... is one more than a multiple of 4, because it is 1 plus some number of 4's. Now 300 is a multiple of 4 itself, so it cannot be one more than a multiple of 4. But 301 is, so 301 is a term.

Filling in the gaps: arithmetic means

Suppose you want to put three numbers between 5 and 25 so that all five make an arithmetic sequence. The 5 numbers have \(5 - 1 = 4\) jumps, and the total climb is \(25 - 5 = 20\), so \(d = 20 \div 4 = 5\): the sequence is 5, 10, 15, 20, 25. The numbers you insert are called arithmetic means.

The simplest case is one number between two: the middle term of three is halfway between its neighbours,

\[\text{if } x,\ y,\ z \text{ is arithmetic, then } y = \frac{x + z}{2}.\]

Why: \(y - x = d\) and \(z - y = d\), so \(y - x = z - y\), which rearranges to \(2y = x + z\). The middle term is the average (the mean) of its two neighbours. That is where the name "arithmetic mean" comes from: the everyday average of two numbers is the number that sits halfway between them.

Worked examples

1. Two terms, going down. An arithmetic sequence has \(t_3 = 20\) and \(t_{10} = -8\). Find \(a\), \(d\) and \(t_n\).

  • Jumps: \(10 - 3 = 7\). Change: \(-8 - 20 = -28\). So \(d = -28 \div 7 = -4\).
  • Back 2 jumps from the 3rd term: \(a = 20 - 2 \times (-4) = 20 + 8 = 28\). (Going back against a negative difference makes the number bigger.)
  • \(t_n = 28 + (n - 1)(-4) = 32 - 4n\). Check: \(t_{10} = 32 - 40 = -8\). ✓

2. Is it a term? Is 500 a term of 7, 13, 19, 25, ...? Is 499?

  • \(a = 7\), \(d = 6\), so \(t_n = 6n + 1\).
  • \(6n + 1 = 500\) gives \(6n = 499\), \(n = 83.1\overline{6}\): not a whole number, so 500 is not a term.
  • \(6n + 1 = 499\) gives \(6n = 498\), \(n = 83\): 499 is the 83rd term.

The quick view: every term is 1 more than a multiple of 6. 499 is \(6 \times 83 + 1\); 500 leaves remainder 2.

3. Inserting means. Put four numbers between 3 and 38 to make an arithmetic sequence.

Six numbers in all, so 5 jumps, and the climb is 35: \(d = 7\). The sequence is 3, 10, 17, 24, 31, 38.

4. A sequence with an unknown inside. The three numbers \(x\), \(2x + 1\), \(4x - 3\) (in that order) make an arithmetic sequence. Find \(x\) and the numbers.

The middle one is the average of the other two, so \(2(2x + 1) = x + (4x - 3)\), that is \(4x + 2 = 5x - 3\), so \(x = 5\). The numbers are 5, 11, 17, with \(d = 6\). ✓

5. Back to the spilled page. We found \(t_n = 4n - 3\), with every term 1 more than a multiple of 4. 300 is not a term; the closest are 297 (the 75th) and 301 (the 76th).

Try it

  1. An arithmetic sequence has \(t_1 = 4\) and \(t_6 = 29\). Find \(d\) and \(t_n\).
  2. An arithmetic sequence has \(t_4 = 2.5\) and \(t_9 = 10\). Find \(a\), \(d\) and \(t_{20}\).
  3. Is 1000 a term of 4, 13, 22, 31, ...? Is 1003?
  4. Put two numbers between 8 and 29 to make an arithmetic sequence.
  5. The numbers \(k + 2\), \(2k\), \(4k - 5\) make an arithmetic sequence in that order. Find \(k\) and the three numbers.
  6. Which numbers are in both 2, 5, 8, 11, ... and 3, 7, 11, 15, ...? Find the first three, and a rule for all of them.
  7. In an arithmetic sequence, the 7th term is three times the 2nd term, and the 10th term is 42. Find the sequence.
  8. Explain why, in any arithmetic sequence, \(t_2 + t_9 = t_4 + t_7\). (Hint: write each one using \(a\) and \(d\).)
Answers
  1. 5 jumps, climb \(29 - 4 = 25\), so \(d = 5\). \(t_n = 4 + (n - 1) \times 5 = 5n - 1\).
  2. 5 jumps, climb \(10 - 2.5 = 7.5\), so \(d = 1.5\). Back 3 jumps: \(a = 2.5 - 4.5 = -2\). \(t_{20} = -2 + 19 \times 1.5 = -2 + 28.5 = 26.5\).
  3. \(a = 4\), \(d = 9\): \(t_n = 9n - 5\). For 1000: \(9n = 1005\), \(n = 111.\overline{6}\), so no. For 1003: \(9n = 1008\), \(n = 112\), so yes, the 112th term. (Every term leaves remainder 4 when divided by 9; 1003 does, 1000 does not.)
  4. Four numbers, 3 jumps, climb 21, so \(d = 7\): 8, 15, 22, 29.
  5. \(2(2k) = (k + 2) + (4k - 5)\), so \(4k = 5k - 3\) and \(k = 3\). The numbers are 5, 6, 7 (difference 1).
  6. Listing: the first goes 2, 5, 8, 11, 14, 17, 20, 23, ... and the second 3, 7, 11, 15, 19, 23, ... The common ones are 11, 23, 35, ... They go up by 12, because a common term must stay common after adding a number that is a multiple of both 3 and 4, and the smallest is 12. The rule is \(12n - 1\).
  7. \(a + 6d = 3(a + d)\) gives \(a + 6d = 3a + 3d\), so \(3d = 2a\). Also \(a + 9d = 42\). From \(3d = 2a\), \(9d = 6a\), so \(a + 6a = 42\), \(a = 6\) and \(d = 4\). The sequence is 6, 10, 14, 18, 22, 26, 30, ... Check: \(t_7 = 30 = 3 \times 10\) ✓ and \(t_{10} = 6 + 36 = 42\) ✓.
  8. \(t_2 + t_9 = (a + d) + (a + 8d) = 2a + 9d\) and \(t_4 + t_7 = (a + 3d) + (a + 6d) = 2a + 9d\). They are equal because the positions add to the same total, \(2 + 9 = 4 + 7 = 11\). This fact is the key to the next lesson.

Common mistake

Dividing by the wrong number of jumps. From the 5th to the 12th term is \(12 - 5 = 7\) jumps. People often count the terms instead (8 of them, from the 5th to the 12th inclusive) and divide by 8. Draw a tiny picture of posts and gaps if you are unsure.

Getting the sign wrong going backwards. To go back from a later term, you subtract \(d\). If \(d\) is negative, subtracting it makes the number bigger: back one step from \(-8\) with \(d = -4\) is \(-8 - (-4) = -4\).

Where it leads