Geometric sequences

​​In a nutshell

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

Finding the $$n^{th}$$ term​

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

$$a, ar, ar^2, ar^3,ar^4,...$$​​

The $$n^{th}$$ term is calculated using a formula involving the first term, $$a$$, and the common ratio, $$r$$. ​

$$\boxed{u_n = ar^{n-1}}$$​​

Example 1

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

Consider the type of sequence this is: 

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

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

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

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Example 2

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

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

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

Rearrange and simplify the equation:​

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

Solve for $$x$$:

$$\begin{aligned} x&= \dfrac{-(-25) \pm \sqrt{(-25)^2 -4(4)(11)}}{2(4)} \\ \\ x&=\dfrac{ 25 \pm \sqrt{449}}{8}\end{aligned}$$​​

For each value of $$x$$ find the common ratio: 

​$$r= \dfrac {\dfrac{ 25 +\sqrt{449}}{8}+1}{\dfrac{ 25 +\sqrt{449}}{8}-4} = \underline{3.82 \ (2 \ d.p.)}$$​​

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​$$r= \dfrac {\dfrac{ 25 -\sqrt{449}}{8}+1}{\dfrac{ 25 -\sqrt{449}}{8}-4} = \underline{-0.42 \ (2 \ d.p.)}$$​​

Example 3

A geometric sequence has a common ratio of $$0.5$$ and its first term is $$10$$. What is the first value in the sequence below $$10^{-2}$$?

Substitute values into the geometric equation and solve for $$n$$​:

$$\begin{aligned} 10^{-2} &= 10 (0.5)^{n-1} \\ 10^{-2} \div 10 &= (0.5)^{n-1} \\ \ln (10^{-3}) &= (n-1)\ln(0.5) \\ \\ \dfrac{ \ln(10^{-3})}{\ln (0.5)}+1& = n \end{aligned}$$​​

​$$n = 10.9657842847$$​

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The common ratio is below $$1$$ therefore, each consecutive term will get closer to zero. This means the first value below  $$10^{-2}$$ will be the $$11^{th}$$ term in the sequence:

$$u_{11} = 10(0.5)^{11-1} \\ \underline{u_{11} = 0.009765625}$$

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