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Explainer Video

Tutor: Labib

Summary

Geometric sequences

​​In a nutshell

Geometric sequences have a common ratio between consecutive terms. This ratio can be used to find the nthn^{th}nth term in a geometric sequence. ​



Finding the nthn^{th}nth term​

A geometric sequence is a sequence such that to go from one term to the next, you multiply be some constant rrr. If the first term is aaa, then the first few terms of the geometric sequence are

a,ar,ar2,ar3,ar4,...a, ar, ar^2, ar^3,ar^4,...a,ar,ar2,ar3,ar4,...​​


The nthn^{th}nth term is calculated using a formula involving the first term, aaa, and the common ratio, rrr. ​

un=arn−1\boxed{u_n = ar^{n-1}}un​=arn−1​​​


Example 1

Given that the first three terms in a sequence are 1.5, 2.25, 3.3751.5, \ 2.25, \ 3.3751.5, 2.25, 3.375, what is the 51st51^{st}51st term in the sequence? ​


Consider the type of sequence this is: 

2.25÷1.5=1.53.375÷2.25=1.5\begin{aligned} 2.25 \div 1.5 &= 1.5 \\ 3.375 \div 2.25 &= 1.5 \end{aligned} 2.25÷1.53.375÷2.25​=1.5=1.5​​​


There is a common ratio, 1.51.51.5, between consecutive terms so this is a geometric sequence. Use the equation for the nthn^{th}nth term, using that a=1.5a=1.5a=1.5 and also r=1.5r=1.5r=1.5​:

u51=(1.5)(1.5)51−1u51=(1.5)(1.5)50u51=(1.5)51u51=956432250.321‾\begin{aligned} u_{51} &= (1.5)(1.5)^{51-1} \\ u_{51}&= (1.5)(1.5)^{50} \\ u_{51} &= (1.5)^{51} \\ \end{aligned} \\ \underline{u_{51} = 956432250.321}u51​u51​u51​​=(1.5)(1.5)51−1=(1.5)(1.5)50=(1.5)51​u51​=956432250.321​​​

​

Example 2

Three consecutive terms of a geometric sequence are x−4, x+1, 5x−3x-4, \ x+1, \ 5x-3x−4, x+1, 5x−3. Given that r≠0r \ne 0r=0, what is the common ratio of this sequence?​


The common ratio between consecutive terms is the same. Use this information to write an equation:

(x+1)(x−4)=(5x−3)(x+1)\dfrac{(x+1)}{(x-4)} = \dfrac{(5x-3)}{(x+1)}(x−4)(x+1)​=(x+1)(5x−3)​​​


Rearrange and simplify the equation:​

(x+1)2=(5x−3)(x−4)x2+2x+1=5x2−23x+120=4x2−25x+11\begin{aligned}(x+1)^2 &= (5x-3)(x-4) \\ x^2 +2x+1 &= 5x^2-23x+12 \\ 0 &= 4x^2 - 25x +11\end{aligned}(x+1)2x2+2x+10​=(5x−3)(x−4)=5x2−23x+12=4x2−25x+11​​​


Solve for xxx:

x=−(−25)±(−25)2−4(4)(11)2(4)x=25±4498\begin{aligned} x&= \dfrac{-(-25) \pm \sqrt{(-25)^2 -4(4)(11)}}{2(4)} \\ \\ x&=\dfrac{ 25 \pm \sqrt{449}}{8}\end{aligned}xx​=2(4)−(−25)±(−25)2−4(4)(11)​​=825±449​​​​​


For each value of xxx find the common ratio: 

​r=25+4498+125+4498−4=3.82 (2 d.p.)‾r= \dfrac {\dfrac{ 25 +\sqrt{449}}{8}+1}{\dfrac{ 25 +\sqrt{449}}{8}-4} = \underline{3.82 \ (2 \ d.p.)}r=825+449​​−4825+449​​+1​=3.82 (2 d.p.)​​​

​

​r=25−4498+125−4498−4=−0.42 (2 d.p.)‾r= \dfrac {\dfrac{ 25 -\sqrt{449}}{8}+1}{\dfrac{ 25 -\sqrt{449}}{8}-4} = \underline{-0.42 \ (2 \ d.p.)}r=825−449​​−4825−449​​+1​=−0.42 (2 d.p.)​​​


Example 3

A geometric sequence has a common ratio of 0.50.50.5 and its first term is 101010. What is the first value in the sequence below 10−210^{-2}10−2?


Substitute values into the geometric equation and solve for nnn​:

10−2=10(0.5)n−110−2÷10=(0.5)n−1ln⁡(10−3)=(n−1)ln⁡(0.5)ln⁡(10−3)ln⁡(0.5)+1=n\begin{aligned} 10^{-2} &= 10 (0.5)^{n-1} \\ 10^{-2} \div 10 &= (0.5)^{n-1} \\ \ln (10^{-3}) &= (n-1)\ln(0.5) \\ \\ \dfrac{ \ln(10^{-3})}{\ln (0.5)}+1& = n \end{aligned} 10−210−2÷10ln(10−3)ln(0.5)ln(10−3)​+1​=10(0.5)n−1=(0.5)n−1=(n−1)ln(0.5)=n​​​

​n=10.9657842847n = 10.9657842847n=10.9657842847​

​

The common ratio is below 111 therefore, each consecutive term will get closer to zero. This means the first value below  10−210^{-2}10−2 will be the 11th11^{th}11th term in the sequence:

u11=10(0.5)11−1u11=0.009765625‾u_{11} = 10(0.5)^{11-1} \\ \underline{u_{11} = 0.009765625}u11​=10(0.5)11−1u11​=0.009765625​


​​


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FAQs - Frequently Asked Questions

How do you find whether a term satisfies conditions in a geometric sequence?

Use logarithms to find if the value of n is an integer.

How do you find any term in a geometric sequence?

By using the calculation ar^{n-1} where a is the first term and r is the common ratio.

What defines a geometric sequence?

There is a common ratio between consecutive terms.

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