Question

A geometric sequence is defined recursively by a(1)=40 and a (n)=a(n-1).1\2 (a) write out the first four terms of this sequence (b) is the 9th term of this sequence larger or smaller than 1\10 show the calculation that you use to determine your answer.

Answer 1

We have the recursive sequence

`\begin{gathered} a_1=40,a_n=(1)/(2)\cdot a_(n-1) \\ \end{gathered}`

A. Calculating the first four terms of the sequence, we get:

`\begin{gathered} a_1=40 \\ a_2=(1)/(2)\cdot40=20 \\ a_3=(1)/(2)\cdot a_2\rightarrow a_3=(1)/(2)\cdot20=10 \\ a_4=(1)/(2)\cdot a_3\rightarrow a_4=(1)/(2)\cdot10=5 \end{gathered}`

This way,

`\begin{gathered} a_1=40 \\ a_2=20 \\ a_3=10 \\ a_4=5 \end{gathered}`

B. Let's calculate the 9th term:

`\begin{gathered} a_5=(1)/(2)\cdot a_4\rightarrow a_5=(1)/(2)\cdot5=(5)/(2) \\ a_6=(1)/(2)\cdot a_5\rightarrow a_6=(1)/(2)\cdot(5)/(2)=(5)/(4) \\ a_7=(1)/(2)\cdot a_6=(1)/(2)\cdot(5)/(4)=(5)/(8) \\ a_8=(1)/(2)\cdot a_7=(1)/(2)\cdot(5)/(8)=(5)/(16) \\ a_9=(1)/(2)\cdot a_8\rightarrow a_9=(1)/(2)\cdot(5)/(16)=(5)/(32) \end{gathered}`

This way, we get that:

`a_9=(5)/(32)`

Now, we need to compare

`\begin{gathered} (1)/(10) \\ \text{and} \\ (5)/(32) \end{gathered}`

To do so, we'll use their LCM: 160

`\begin{gathered} (1)/(10)\rightarrow(16)/(160) \\ \text{and} \\ (5)/(32)\rightarrow(25)/(160) \end{gathered}`

Therefore, we can conclude that

`a_9>(1)/(10)`

(The term is larger than 1/10)