Math Detours · Grade 9 · Sharpen · 11

The nth term

Jump straight to the hundredth number of a sequence without writing the first ninety-nine. Constant difference gives a linear rule, constant ratio an exponential one.

Name ____________________Date ____________

Tonight's idea

A sequence is a list of numbers in order. Write \(a_n\) for the term in position \(n\). The small \(n\) is a label, not a power: \(a_1\) is the first term, \(a_4\) the fourth. A rule for \(a_n\) lets you jump to any term.

Take 5, 8, 11, 14, … Each term is 3 more than the last: a constant difference \(d = 3\).

+3+3+3 581114? n = 1n = 2n = 3n = 4n
Four terms, three jumps. Reaching term \(n\) from term 1 takes \(n - 1\) jumps, not \(n\).

Constant difference \(d\): \(\;a_n = a_1 + (n - 1)d\).

Here \(a_n = 5 + 3(n - 1) = 3n + 2\). That is a linear rule, with \(d\) playing the part of the slope in \(y = mx + b\).

Now take 5, 10, 20, 40, … Each term is 2 times the last: a constant ratio \(r = 2\). It is still \(n - 1\) jumps, but each jump multiplies.

Constant ratio \(r\): \(\;a_n = a_1 \cdot r^{\,n-1}\).

Here \(a_n = 5 \cdot 2^{\,n-1}\), an exponential rule. Check \(n = 4\): \(5 \times 2^3 = 40\).

Watch one: the 100th term of 7, 11, 15, 19, …

The differences are all 4, so \(d = 4\) and \(a_1 = 7\).

\(a_n = 7 + 4(n - 1) = 4n + 3\). Check \(n = 2\): \(8 + 3 = 11\). Then \(a_{100} = 403\), with no need for the 99 terms before it.

Is 201 in the sequence? Solve \(4n + 3 = 201\): \(n = 49.5\). A position must be a whole number, so no. But 203 is, at \(n = 50\).

Next year you add the terms up as well: Sequences, series, and Gauss's trick.

Your turn

  1. Find a rule for \(a_n\), then the term asked for. (a) 4, 9, 14, 19, …; the 50th term. (b) 20, 17, 14, 11, …; the 50th term. (c) 2, 6, 18, 54, …; the 8th term.
  2. A sequence has \(a_n = 6n - 5\). (a) Write its first four terms. (b) Which term is 295? (c) Is 500 a term? Say how you know.
  3. A row of squares is built from matchsticks, each square sharing a side with the next. One square needs 4 sticks, two need 7, three need 10. (a) Find a rule for the sticks in a row of \(n\) squares, and say where each part of the rule comes from. (b) How many complete squares can you make with 100 sticks?
  4. (a) A sequence with a constant difference has 3rd term 17 and 7th term 41. Find \(d\), then \(a_1\), then \(a_{100}\). (b) A sequence with a constant positive ratio has 3rd term 12 and 5th term 48. Find \(r\), \(a_1\) and a rule for \(a_n\).

Stretch: one sequence inside another

Both 3, 6, 9, 12, … and 3, 6, 12, 24, … start 3, 6. Explain why every term of the second appears somewhere in the first. At which position in the first does the 10th term of the second sit?

Answers

  1. (a) \(d = 5\): \(a_n = 4 + 5(n - 1) = 5n - 1\), and \(a_{50} = 249\). (b) \(d = -3\): \(a_n = 20 - 3(n - 1) = 23 - 3n\), and \(a_{50} = -127\). (c) \(r = 3\): \(a_n = 2 \cdot 3^{\,n-1}\), and \(a_8 = 2 \times 3^7 = 4374\).
  2. (a) 1, 7, 13, 19. (b) \(6n - 5 = 295\) gives \(n = 50\): the 50th. (c) \(6n - 5 = 500\) gives \(n = 505/6\), not a whole number, so no.
  3. (a) \(a_n = 4 + 3(n - 1) = 3n + 1\). Each square adds 3 new sticks; the 1 is the stick that closes the left end. (b) \(3n + 1 \le 100\) gives \(n \le 33\): 33 squares, using all 100 sticks.
  4. (a) From the 3rd term to the 7th is 4 jumps: \(4d = 41 - 17 = 24\), so \(d = 6\). Two jumps back from 17: \(a_1 = 5\). \(a_n = 6n - 1\), so \(a_{100} = 599\). (b) Two jumps: \(r^2 = 48/12 = 4\), so \(r = 2\). Two jumps back from 12: \(a_1 = 12/4 = 3\). \(a_n = 3 \cdot 2^{\,n-1}\).

Stretch. The first sequence is \(3n\): every positive multiple of 3. The second is \(3 \cdot 2^{\,n-1}\), always 3 times a whole number, so each term is a multiple of 3 and is in the first. Its 10th term is \(3 \times 2^9 = 1536 = 3 \times 512\), which sits at position 512.

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