Harder geometric progressions · Sequences, series & binomial expansion

The G.P. formulas from General Maths (see sequences and series) are all you need here:

nth term of a G.P.Tₙ = arⁿ⁻¹

Sum to infinityS∞ = a ÷ (1 − r), when −1 < r < 1

Sum of n termsSₙ = a(rⁿ − 1) ÷ (r − 1) = a(1 − rⁿ) ÷ (1 − r)

As with an A.P., each fact becomes an equation in $a$ and $r$ . The difference is how you solve them: G.P. terms are built by multiplying by $r$ , so you usually divide one equation by the other.

Two facts: divide to remove a

When both equations have $a$ as a factor, dividing one by the other cancels it, leaving an equation in $r$ alone. Factorise first if the facts are sums of terms.

Two facts: divide to remove aStep through

ar² = 12 and ar⁵ = 96

workingstep 0 of 3

Write each term as arⁿ⁻¹: the 3rd term is ar², the 6th is ar⁵.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q3

The sum of the second and third terms of a Geometric Progression (G.P.) is 48. If the sum of the third and fourth terms is 144, find the first term of the progression.

Write the facts
- ${ar + ar^2 = 48}$ , which factorises to ${ar(1 + r) = 48}$ .
- ${ar^2 + ar^3 = 144}$ , which factorises to ${ar^2(1 + r) = 144}$ .
Think first. Write the 2nd, 3rd and 4th terms with a and r.
Divide
- ${\frac{ar^2(1 + r)}{ar(1 + r)} = \frac{144}{48}}$ .
- $a$ , one $r$ and the bracket cancel: ${r = 3}$ .
Think first. Divide the second equation by the first. What cancels?
Find a
- ${a(3)(4) = 48}$ , so ${12a = 48}$ .
- So the first term is ${a = 4}$ .
Think first. Put r = 3 into ar(1 + r) = 48.

More: two facts

Three terms: write them as a/r, a, ar

When a question gives the product of three consecutive terms, call the middle term $a$ . The terms are then $\frac ar$ , $a$ and $ar$ , and the $r$ ‘s cancel in the product:

Three consecutive termsThe product is a³, so a comes out at once

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q11 (a)

The sum of the first three terms of a decreasing exponential sequence (G.P.) is equal to 7 and the product of these three terms is equal to 8. Find the: (i) common ratio; (ii) first three terms of the sequence.

Use the product
- ${\frac ar \times a \times ar = a^3 = 8}$ .
- So ${a = 2}$ : the middle term is 2.
Think first. The terms are a/r, a, ar. What is their product?
Use the sum
- ${\frac2r + 2 + 2r = 7}$ .
- Multiply by $r$ : ${2 + 2r + 2r^2 = 7r}$ .
- Rearrange: ${2r^2 - 5r + 2 = 0}$ .
- Factorise: ${(2r - 1)(r - 2) = 0}$ , so ${r = \frac12}$ or ${r = 2}$ .
Think first. Write 2/r + 2 + 2r = 7 and clear the fraction.
Choose r
- A decreasing G.P. with positive terms has ${r < 1}$ , so ${r = \frac12}$ .
- The terms are ${\frac{2}{1/2} = 4}$ , then $2$ , then ${2 \times \frac12 = 1}$ : that is $4, 2, 1$ .
Think first. The sequence is decreasing. Which r?

More: three terms

WAEC 2022 · Paper 1 · Q26The first, second and third terms of an exponential sequence (G.P.) are $(x - 4)$ , $(x + 2)$ and $(3x + 1)$ respectively. …

The sum to infinity with another fact

$S_\infty = \frac{a}{1 - r}$ is one equation. With a second fact, make $a$ the subject of one and substitute into the other. Look out for $(1 - r)(1 + r) = 1 - r^2$ .

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q11

The sum to infinity of an exponential sequence (G.P.) with a positive common ratio is 25 and the sum of the first two terms is 16. Find the:

fifth term;

sum of the first four terms.

Two equations
- ${\frac{a}{1 - r} = 25}$ , so ${a = 25(1 - r)}$ .
- ${a + ar = 16}$ , so ${a(1 + r) = 16}$ .
Think first. Write S∞ = 25 and T₁ + T₂ = 16.
Find r
- ${25(1 - r)(1 + r) = 16}$ .
- So ${1 - r^2 = \frac{16}{25}}$ , and ${r^2 = \frac{9}{25}}$ .
- The ratio is positive, so ${r = \frac35}$ .
- Then ${a = 25\left(1 - \frac35\right) = 10}$ .
Think first. Substitute a = 25(1 − r) into a(1 + r) = 16.
The fifth term
- ${T_5 = 10\left(\frac35\right)^4 = 10 \times \frac{81}{625}}$ .
- ${= \frac{810}{625} = \frac{162}{125}}$ .
Think first. T₅ = ar⁴.
The sum of four terms
- ${S_4 = \frac{a(1 - r^4)}{1 - r} = 25\left(1 - \frac{81}{625}\right)}$ .
- ${= 25 \times \frac{544}{625} = \frac{544}{25} = 21\frac{19}{25}}$ .

More: sums of a G.P.

An A.P. and a G.P. together

When terms of an A.P. form a G.P., write them with $a$ and $d$ , then use the G.P. condition for three consecutive terms: the middle one squared equals the product of the other two.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q11 (b)

The second, fourth and eighth terms of an Arithmetic Progression (A.P.) form the first three consecutive terms of a Geometric Progression (G.P.). The sum of the third and fifth terms of the A.P. is 20. Find the: (i) first four terms of the A.P.; (ii) sum of the first ten terms of the A.P.

The A.P. terms
- ${T_2 = a + d}$ , ${T_4 = a + 3d}$ and ${T_8 = a + 7d}$ .
Think first. Write the 2nd, 4th and 8th terms with a and d.
The G.P. condition
- ${(a + 3d)^2 = (a + d)(a + 7d)}$ .
- Expand: ${a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2}$ .
- Simplify: ${2d^2 = 2ad}$ , so ${d(d - a) = 0}$ .
- An A.P. with $d = 0$ would be constant, so ${d = a}$ .
Think first. Middle squared = product of the other two.
The second fact
- ${(a + 2d) + (a + 4d) = 20}$ , so ${2a + 6d = 20}$ .
- With ${d = a}$ : ${8a = 20}$ , so ${a = d = \frac52}$ .
- The first four terms: ${\frac52, 5, \frac{15}{2}, 10}$ .
Think first. T₃ + T₅ = 20. Write it with a and d.
The sum of ten terms
- ${S_{10} = \frac{10}{2}\left[2\left(\frac52\right) + 9\left(\frac52\right)\right]}$ .
- ${= 5\left(5 + \frac{45}{2}\right) = 5 \times \frac{55}{2} = 137.5}$ .

Your turn

WAEC 2022 · Paper 2 · Q9 (b)

(b)
The sum of the first and third terms of a Geometric Progression (G.P.) is 20 and the product of the first and fourth terms is 18 times the second term. If the common ratio of the G.P. is positive, find the sum of the first 8 terms.

Worked solution (try it first)

(b)

Let the first term be $a$ and the common ratio $r$ .
The first and third terms add up to 20: $a + ar^2 = 20$ .
The product of the first and fourth terms is 18 times the second: $a \times ar^3 = 18ar$ .
Divide both sides by $ar$ : $ar^2 = 18$ .
Substitute into the first equation: $a + 18 = 20$ , so $a = 2$ .
Then $2r^2 = 18$ , so $r^2 = 9$ and $r = 3$ (the ratio is positive).
$S_8 = \dfrac{a(r^8 - 1)}{r - 1}$
$= \dfrac{2(6561 - 1)}{2}$
$= 6560$ .

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