Sequences & series (AP, GP) · Lesson 3 of 4

Geometric progressions

Geometric progressions and the formula Tₙ = arⁿ⁻¹, finding a and r from two terms by dividing, the two ratios an even power gives, and three consecutive terms.

15 minYou should already know: Expressions, formulae & change of subject
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In a geometric progression (G.P., or exponential sequence) you multiply by the same number every time. That number is the common ratio rr, found by dividing any term by the one before: r=T2÷T1r = T_2 \div T_1. With first term aa:

a,ar,ar2,ar3, …a,\quad ar,\quad ar^2,\quad ar^3,\ \dots

The first term has no rr, the second has one, the third has r2r^2. So the nnth term has rr to the power n−1n - 1:

Tn=arn−1T_n = ar^{n - 1}
aar×rar²×rar³×r
nth term of a G.P.Tₙ = arⁿ⁻¹: each bar is r times the one before

Compare an A.P., where you add dd each time and Tn=a+(n−1)dT_n = a + (n - 1)d. To tell them apart, check the differences and the ratios of the terms: 2,6,18,542, 6, 18, 54 has differences 4,12,364, 12, 36 (not equal) but ratios 3,3,33, 3, 3, so it’s a G.P.

Try it

The terms of a G.P.Set a, pick r, then pick the terms
013263124245486967192
48T53 × 2⁴arⁿ⁻¹4times multiplied by r
Each term is bigger than the last: the terms grow faster and faster. T5 is a = 3 multiplied by r = 2 4 times, one fewer than n: T5 = 3 × 2⁴ = 48.

Try each ratio. With r=2r = 2 or 33 the terms grow faster and faster; with r=12r = \frac12 they shrink towards 0; with a negative rr the signs alternate. In “nth term” mode the power on rr is always one less than nn.

Finding a and r from two terms

Write each term as arn−1ar^{n - 1}, then divide one equation by the other: aa cancels and leaves a power of rr. From TpT_p to TqT_q you multiply by rr (q−p)(q - p) times, so Tq÷Tp=rq−pT_q \div T_p = r^{q - p}.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q11 (a)

The third and sixth terms of a Geometric Progression (G.P.) are 14\frac14 and 132\frac{1}{32} respectively. Find the: (i) first term and the common ratio; (ii) seventh term.

  1. Write the two terms

    T3=ar2=14T_3 = ar^2 = \frac14 and T6=ar5=132T_6 = ar^5 = \frac{1}{32}.

    Think first. Write the 3rd and 6th terms using a and r.

  2. Divide to find r

    ar5ar2=r3=132÷14=432=18\dfrac{ar^5}{ar^2} = r^3 = \dfrac{1}{32} \div \dfrac14 = \dfrac{4}{32} = \dfrac18, so r=12r = \frac12.

    Think first. Divide T₆ by T₃. What happens to a?

  3. (i) The first term

    a×14=14a \times \frac14 = \frac14, so a=1a = 1. The first term is 1 and the common ratio is 12\frac12.

    Think first. Put r = ½ into ar² = ¼.

  4. (ii) The seventh term

    T7=ar6=1×(12)6=164T_7 = ar^6 = 1 \times \left(\frac12\right)^6 = \frac{1}{64}.

    Think first. T₇ has r to which power?

When the power is even

If dividing leaves an even power, such as r2=4r^2 = 4, there are two answers: r=2r = 2 or r=−2r = -2. Both give the same even powers. Keep both unless the question says the ratio is positive, or the terms rule one out.

Three consecutive terms

If xx, yy and zz are consecutive terms of a G.P., the ratios are equal: yx=zy\dfrac{y}{x} = \dfrac{z}{y}. Cross-multiply:

y2=xzy^2 = xz

The middle term squared equals the product of its neighbours. (yy is the geometric mean of xx and zz.)

xyz×r×r
Consecutive terms of a G.P.y ÷ x = z ÷ y, so y² = xz

Worked example · WAEC 2024

WAEC 2024 · Paper 2 · Q6

Given that (x+2)(x + 2), (4x+3)(4x + 3) and (7x+24)(7x + 24) are consecutive terms of a geometric progression (G.P.), find the:

values of xx;

common ratio (for each value of xx).

  1. Set the ratios equal

    4x+3x+2=7x+244x+3\dfrac{4x + 3}{x + 2} = \dfrac{7x + 24}{4x + 3}, so (4x+3)2=(x+2)(7x+24)(4x + 3)^2 = (x + 2)(7x + 24).

    Think first. Which equation links three consecutive terms of a G.P.?

  2. Expand both sides

    16x2+24x+9=7x2+38x+4816x^2 + 24x + 9 = 7x^2 + 38x + 48

    Think first. Square the left side; multiply out the right.

  3. (a) Solve

    9x2−14x−39=09x^2 - 14x - 39 = 0, which factorises as (x−3)(9x+13)=0(x - 3)(9x + 13) = 0. So x=3x = 3 or x=−139x = -\frac{13}{9}.

    Think first. Collect everything on one side and factorise.

  4. (b) The ratio when x = 3

    The terms are 5,15,455, 15, 45, so r=155=3r = \frac{15}{5} = 3.

    Think first. Work out the three terms first.

  5. (b) The ratio when x = −13/9

    x+2=59x + 2 = \frac59 and 4x+3=−529+279=−2594x + 3 = -\frac{52}{9} + \frac{27}{9} = -\frac{25}{9}, so r=−259÷59=−5r = -\frac{25}{9} \div \frac59 = -5. (The third term, 1259\frac{125}{9}, is −5-5 times −259-\frac{25}{9} ✓.)

    Think first. Work out x + 2 and 4x + 3 as fractions.

Your turn

WAEC 2014 · Paper 2 · Q6 (b)

  1. (b)

    The sum of the first and third terms of a Geometric Progression (G.P.) is 40 while the fourth and sixth terms are in the ratio 1:41 : 4. Find the: (i) common ratio; (ii) fifth term.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(b)(i)

  1. With first term aa and common ratio rr, the terms are a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \ldots The fourth and sixth terms are in the ratio 1:41 : 4: ar3ar5=14\frac{ar^3}{ar^5} = \frac14, so 1r2=14\frac1{r^2} = \frac14 and r2=4r^2 = 4.
  2. So r=2r = 2 or r=−2r = -2.

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