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Finding a and r

Patterns and sequences · level 6, lesson 3 of 6

A question to start

A geometric sequence has been half rubbed off a whiteboard. Only two terms are left:

the 3rd term is 4, and the 5th term is 36.

What was the sequence? Try it before reading on. Most people find one answer. There are actually two different geometric sequences that fit, and both are correct. By the end of this lesson you will see where the second one hides, and when it can and cannot happen.

The idea

From two terms: divide, do not subtract

In level 5 you found \(a\) and \(d\) of an arithmetic sequence from two terms by subtracting: the difference between the terms is \(d\) times the number of steps. For a geometric sequence everything moves "one floor up":

  • each step multiplies by \(r\), so going several steps multiplies by a power of \(r\);
  • so you divide the terms instead of subtracting them, and take a root instead of dividing.

Try it with \(t_2 = 12\) and \(t_5 = 96\). From position 2 to position 5 is 3 steps:

× r × r × r t₂ = 12 t₅ = 96 12 × r × r × r = 96, so r³ = 8
\[r^3 = \frac{t_5}{t_2} = \frac{96}{12} = 8, \qquad\text{so}\qquad r = 2,\]

because \(2 \times 2 \times 2 = 8\). Then step back once from the 2nd term: \(a = t_2 \div r = 12 \div 2 = 6\). The sequence is 6, 12, 24, 48, 96, ... and \(t_n = 6 \times 2^{n-1}\).

In general, for terms \(t_m\) and \(t_k\) with \(k > m\):

\[r^{\,k - m} = \frac{t_k}{t_m}.\]

With algebra, the two facts are \(ar = 12\) and \(ar^4 = 96\). Dividing the second by the first cancels \(a\) and leaves \(r^3 = 8\): the same calculation. Dividing removes the shared start \(a\), just as subtracting did for arithmetic sequences.

The hidden second answer: even numbers of steps

Now the whiteboard: \(t_3 = 4\) and \(t_5 = 36\). Two steps, so

\[r^2 = \frac{36}{4} = 9.\]

Which numbers square to 9? Both \(3\) and \(-3\), because \((-3) \times (-3) = 9\) too. So there are two sequences:

  • \(r = 3\): \(a = t_3 \div r^2 = 4 \div 9 = \tfrac49\), giving \(\tfrac49, \tfrac43, \mathbf{4}, 12, \mathbf{36}, 108, \ldots\)
  • \(r = -3\): \(a = 4 \div 9 = \tfrac49\) again (since \(r^2 = 9\) either way), giving \(\tfrac49, -\tfrac43, \mathbf{4}, -12, \mathbf{36}, -108, \ldots\)

Both have 4 in position 3 and 36 in position 5. The terms in between, and the terms in even positions, differ in sign. Whenever the number of steps between the two known terms is even, look for a positive and a negative ratio.

When the number of steps is odd, there is only one answer, and its sign is forced. For example, \(t_1 = 2\) and \(t_4 = -54\) give \(r^3 = -27\), so \(r = -3\): a negative number cubed is negative, and no positive number cubed is \(-27\).

And sometimes there is no answer: if \(t_1 = 2\) and \(t_3 = -18\), then \(r^2 = -9\), but no real number squares to a negative number. These two terms cannot belong to the same geometric sequence (two terms an even number of steps apart always have the same sign).

The middle of three: the geometric mean

If \(x\), \(y\), \(z\) are three terms in a row of a geometric sequence, then \(\dfrac{y}{x} = \dfrac{z}{y}\) (both are \(r\)). Multiply both sides by \(xy\):

\[y^2 = xz.\]

The middle term squared equals the product of its neighbours. The positive number whose square is \(xz\), that is \(\sqrt{xz}\), is called the geometric mean of \(x\) and \(z\). Compare the arithmetic mean \(\dfrac{x + z}{2}\) from level 5. For 4 and 25: the arithmetic mean is 14.5 (halfway by adding), while the geometric mean is \(\sqrt{100} = 10\) ("halfway" by multiplying: \(4 \times 2.5 = 10\) and \(10 \times 2.5 = 25\)).

Arithmetic, geometric or neither?

To classify a sequence, run both tests:

  1. Differences (later minus earlier). All equal: arithmetic.
  2. Ratios (later divided by earlier). All equal: geometric.
  3. Neither all equal: neither. (Then lesson 4 of level 5 may still help: maybe the second differences are constant, and it is quadratic.)

A sequence can be both only if it is constant and not zero, like 5, 5, 5: difference 0 and ratio 1. Any sequence containing 0 (after the first term) is not geometric.

You can also classify from the rule. \(t_n = 3n + 2\) is arithmetic (\(n\) is multiplied by a number). \(t_n = 5 \times 2^n\) is geometric (\(n\) is an exponent): each step multiplies by 2. \(t_n = n^2\) is neither (\(n\) is the base, not the exponent). And \(t_n = 2^n + 1\) is neither: the ratios are \(\tfrac53, \tfrac95, \tfrac{17}{9}, \ldots\) The "+1" spoils the multiplying.

Worked examples

1. A shrinking sequence. \(t_3 = 100\) and \(t_6 = 12.5\). Find \(a\) and \(r\).

\(r^3 = 12.5 \div 100 = 0.125 = \tfrac18\), so \(r = \tfrac12\). Back two steps: \(a = 100 \div \left(\tfrac12\right)^2 = 100 \times 4 = 400\). The sequence is 400, 200, 100, 50, 25, 12.5, ...

2. Two answers. \(t_2 = 6\) and \(t_4 = 54\). Find all possible sequences.

\(r^2 = 54 \div 6 = 9\), so \(r = 3\) or \(r = -3\). Then \(a = 6 \div r\):

  • \(r = 3\), \(a = 2\): 2, 6, 18, 54, 162, ...
  • \(r = -3\), \(a = -2\): −2, 6, −18, 54, −162, ...

3. Inserting geometric means. Put two numbers between 4 and 108 so that all four make a geometric sequence.

Four terms, 3 steps: \(r^3 = 108 \div 4 = 27\), so \(r = 3\). The sequence is 4, 12, 36, 108.

4. An unknown inside. The numbers \(x\), \(x + 3\), \(x + 9\) form a geometric sequence. Find \(x\).

Middle squared equals the product of the neighbours: \((x + 3)^2 = x(x + 9)\). Multiply out: \(x^2 + 6x + 9 = x^2 + 9x\). The \(x^2\) cancels, leaving \(9 = 3x\), so \(x = 3\). The sequence is 3, 6, 12, ratio 2. ✓

5. Classify.

Sequence Differences Ratios Kind
3, 9, 27, 81 6, 18, 54 3, 3, 3 geometric
3, 9, 15, 21 6, 6, 6 3, 1.67, 1.4 arithmetic
3, 9, 18, 30 6, 9, 12 3, 2, 1.67 neither (quadratic)
81, −27, 9, −3 −108, 36, −12 \(-\tfrac13\), \(-\tfrac13\), \(-\tfrac13\) geometric

Try it

  1. A geometric sequence has \(t_1 = 5\) and \(t_4 = 40\). Find \(r\) and \(t_n\).
  2. A geometric sequence has \(t_2 = 10\) and \(t_4 = 250\). Find both possible sequences (first four terms each).
  3. A geometric sequence has \(t_4 = 2\) and \(t_7 = -\tfrac{1}{4}\). Find \(r\) and \(a\).
  4. Classify each as arithmetic, geometric, both or neither: (a) 2, 2, 2, 2; (b) 1, 2, 4, 7, 11; (c) 10, 1, 0.1, 0.01; (d) 0, 3, 6, 9; (e) 0, 3, 9, 27.
  5. Find the geometric mean of 2 and 32, and the arithmetic mean. Which is bigger?
  6. The numbers \(x\), \(x + 4\), \(x + 12\) form a geometric sequence. Find \(x\) and the ratio.
  7. Is 1458 a term of 2, 6, 18, 54, ...? Is 1000?
  8. A sequence has \(t_n = 3 \times 2^n\). Write its first three terms and give \(a\) and \(r\). Why is \(a\) not 3?
Answers
  1. 3 steps: \(r^3 = 40 \div 5 = 8\), so \(r = 2\) and \(t_n = 5 \times 2^{n-1}\).
  2. \(r^2 = 250 \div 10 = 25\), so \(r = 5\) or \(r = -5\). With \(r = 5\): \(a = 2\), giving 2, 10, 50, 250. With \(r = -5\): \(a = -2\), giving −2, 10, −50, 250.
  3. 3 steps: \(r^3 = -\tfrac14 \div 2 = -\tfrac18\), so \(r = -\tfrac12\) (odd number of steps: only one answer). Back 3 steps: \(a = 2 \div \left(-\tfrac12\right)^3 = 2 \div \left(-\tfrac18\right) = -16\). Check: −16, 8, −4, 2, −1, \(\tfrac12\), \(-\tfrac14\). ✓
  4. (a) Both: difference 0, ratio 1. (b) Differences 1, 2, 3, 4: neither (but quadratic, second difference 1). (c) Ratios all 0.1: geometric. (d) Differences all 3: arithmetic. (e) Not arithmetic (differences 3, 6, 18) and not geometric (a zero term, and \(3 \div 0\) is undefined): neither.
  5. Geometric mean \(\sqrt{2 \times 32} = \sqrt{64} = 8\) (and 2, 8, 32 has ratio 4). Arithmetic mean \(\dfrac{2 + 32}{2} = 17\). The arithmetic mean is bigger. (For two different positive numbers it always is; this is a famous inequality you can prove later.)
  6. \((x + 4)^2 = x(x + 12)\): \(x^2 + 8x + 16 = x^2 + 12x\), so \(16 = 4x\) and \(x = 4\). The sequence is 4, 8, 16 with ratio 2.
  7. \(t_n = 2 \times 3^{n-1}\). For 1458: \(3^{n-1} = 729 = 3^6\), so \(n = 7\): yes, the 7th term. For 1000: \(3^{n-1} = 500\), but \(3^5 = 243\) and \(3^6 = 729\), so no power of 3 equals 500: not a term.
  8. \(t_1 = 3 \times 2 = 6\), \(t_2 = 12\), \(t_3 = 24\). So \(a = 6\) and \(r = 2\). The formula has \(2^n\), not \(2^{n-1}\), so even the first term has one factor of 2 in it: \(3 \times 2^n = 6 \times 2^{n-1}\).

Common mistake

Forgetting the negative ratio. If \(r^2 = 9\) you must consider \(r = -3\) as well as \(r = 3\). It happens whenever the two known terms are an even number of steps apart.

Subtracting instead of dividing. For \(t_2 = 12\) and \(t_5 = 96\), the difference \(96 - 12 = 84\) means nothing for a geometric sequence. Divide: \(96 \div 12 = 8 = r^3\).

Where it leads