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iGCSE (2021 Edition)

9.03 Geometric sequences

Lesson

 

A sequence in which each term increases or decreases from the last by a constant factor is called a geometric sequence. We refer to the constant factor the terms are changing by as the common ratio, which will result from dividing any two successive terms $\left(\frac{u_{n+1}}{u_n}\right)$(un+1un).

We denote the first term in the sequence by the letter $u_1$u1 , or $a,$a, and the common ratio by the letter $r$r. For example, the sequence $4,8,16,32\dots$4,8,16,32 is geometric with $u_1=4$u1=4 and $r=2$r=2. The sequence $100,-50,25,-12.5,\dots$100,50,25,12.5, is geometric with $u_1=100$u1=100 and $r=\frac{-1}{2}$r=12.

To describe the rule in words we say "next term is $r$r multiplied by previous term". 

We can find an explicit formula in terms of $u_1$u1 and $r$r, this is useful for finding the $n$nth term without listing the sequence.

Let's look at a table of a concrete example to see the pattern for the explicit formula. For the sequence $5,10,20,40,\ldots$5,10,20,40,, we have a starting term of $5$5 and a common ratio of $2$2, that is $u_1=5$u1=5 and $r=2$r=2. A table of the sequence is show below:

$n$n $u_n$un Pattern
$1$1 $5$5 $5\times2^0$5×20
$2$2 $10$10 $5\times2^1$5×21
$3$3 $20$20 $5\times2^2$5×22
$4$4 $40$40 $5\times2^3$5×23
...    
$n$n $u_n$un $5\times2^{n-1}$5×2n1

 

The pattern starts to become clear and we could guess that the tenth term is $u_{10}=5\times2^9$u10=5×29  and the one-hundredth term is $u_{100}=5\times2^{99}$u100=5×299. And following the pattern, the explicit formula for the $n$nth term is $u_n=5\times2^{n-1}$un=5×2n1.

For any geometric progression with starting value $u_1$u1 and common ratio $r$r has the terms given by: $u_1,u_1r,u_1r^2,u_1r^3,...$u1,u1r,u1r2,u1r3,... We see a similar pattern to our previous table and can write down the formula for the $n$nth term:

$u_n=u_1r^{n-1}$un=u1rn1

 

Forms of geometric sequences

For any geometric sequence with starting value $a$a and common ratio $r$r, we can express it in the form:

$t_n=ar^{n-1}$tn=arn1

 

Worked examples

Example 1

For the sequence  $15,30,60,120...$15,30,60,120..., find an explicit rule for the $n$nth term and hence, find the $8$8th term. 

Think: Check that the sequence is geometric, is the next term made by multiplying the previous term by a constant factor? Then write down the starting value $u_1$u1 and common ratio $r$r and substitute these into the general form: $u_n=u_1r^{n-1}$un=u1rn1

Do: Dividing the second term by the first we get, $\frac{u_2}{u_1}=\frac{30}{15}=\frac{2}{1}$u2u1=3015=21. Checking the ratio between the successive pairs we also get $\frac{2}{1}$21.  So we have a geometric sequence with: $u_1=15$u1=15 and $r=2$r=2. The general formula for this sequence is: $u_n=15\left(2\right)^{n-1}$un=15(2)n1.

Hence, the $8$8th term is:

$u_8$u8 $=$= $15\left(2\right)^7$15(2)7
  $=$= $15\times128$15×128
  $=$= $1920$1920

 

 

Practice questions

Question 1

Study the pattern for the following sequence, and write down the next two terms.

  1. $3$3, $15$15, $75$75, $\editable{}$, $\editable{}$

Question 2

Consider the first four terms in this geometric sequence: $-8$8, $-16$16, $-32$32, $-64$64

  1. If $T_n$Tn is the $n$nth term, evaluate $\frac{T_2}{T_1}$T2T1.

  2. Evaluate $\frac{T_3}{T_2}$T3T2

  3. Evaluate $\frac{T_4}{T_3}$T4T3

  4. Hence find the value of $T_5$T5.

Question 3

In a geometric progression, $T_4=54$T4=54 and $T_6=486$T6=486.

  1. Solve for $r$r, the common ratio in the sequence. Write both solutions on the same line separated by a comma.

  2. For the case where $r=3$r=3, solve for $a$a, the first term in the progression.

  3. Consider the sequence in which the first term is positive. Find an expression for $T_n$Tn, the general $n$nth term of this sequence.

Outcomes

0606C12.3B

Recognise geometric progressions.

0606C12.4B

Use the formulae for the nth term and for the sum of the first n terms to solve problems involving geometric progressions.

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