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iGCSE (2021 Edition)

13.09 Geometric sequences (Extended)

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 $$(un+1un).

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

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

We can find an explicit formula in terms of $$u1 and $$r, this is useful for finding the $$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 $$5,10,20,40,, we have a starting term of $$5 and a common ratio of $$2, that is $$u1=5 and $$r=2. A table of the sequence is show below:

$$n $$un Pattern
$$1 $$5 $$5×20
$$2 $$10 $$5×21
$$3 $$20 $$5×22
$$4 $$40 $$5×23
...    
$$n $$un $$5×2n1

 

The pattern starts to become clear and we could guess that the tenth term is $$u10=5×29  and the one-hundredth term is $$u100=5×299. And following the pattern, the explicit formula for the $$nth term is $$un=5×2n1.

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

$$un=u1rn1

 

Forms of geometric sequences

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

$$tn=arn1

 

Worked examples

Example 1

For the sequence  $$15,30,60,120..., find an explicit rule for the $$nth term and hence, find the $$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 $$u1 and common ratio $$r and substitute these into the general form: $$un=u1rn1

Do: Dividing the second term by the first we get, $$u2u1=3015=21. Checking the ratio between the successive pairs we also get $$21.  So we have a geometric sequence with: $$u1=15 and $$r=2. The general formula for this sequence is: $$un=15(2)n1.

Hence, the $$8th term is:

$$u8 $$= $$15(2)7
  $$= $$15×128
  $$= $$1920

 

 

Practice questions

Question 1

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

  1. $$3, $$15, $$75, $$, $$

Question 2

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

  1. If $$Tn is the $$nth term, evaluate $$T2T1.

  2. Evaluate $$T3T2

  3. Evaluate $$T4T3

  4. Hence find the value of $$T5.

Question 3

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

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

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

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

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