Question

help me with this please (d) Write a definition for the nth term of sequence B(e) If these sequences continue, then which is greater, A(9) or B(9)? Explain or show how you know.

Answer 1

Step-by-step explanation:

We are given a sequence of numbers and each number as shown in the table labeled Sequence A is derived by multiplying the previous number by 2.

That is;

`0th=(1)/(4);1st=(1)/(4)*2;2nd=(1)/(2)*2;3rd=1*2;4th=2*2`

We can observe that the common ratio is 2, for every new term.

What we have is a geometric sequence and the variables are as follows;

`a_1=(1)/(2);r=2`

If we use the term number 1 as the first term (that is 1/2), then the formula for this sequence will be;

`a_n=ar^(n-1)`

For the second term we observe that each next term is derived by adding 10 to the previous one. Hence we have;

`0th=2;1st=\left(2+10\right);2nd=\left(12+10\right);3rd=\left(22+10\right);4th=\left(32+10\right)`

The common difference here is 10, and we can see that sequence B is an arithmetic sequence. Using term 1 as the first term, and the common difference as 10, the nth term for this sequence will be;

`\begin{gathered} a_n=a+\left(n-1\right)d \\ Note\text{ that;} \\ a_1=12;d=10 \end{gathered}`

To determine A(9), that is the 9th term for sequence A;

`a_n=ar^(n-1)`
`a_9=(1)/(2)\left(2\right)^(9-1)`
`a_9=(1)/(2)\left(2\right)^8`
`a_9=(1)/(2)\left(256\right)`
`a_9=128`

To determine B(9), that is the 9th term for sequence B;

`a_n=a+\left(n-1\right)d`
`a_9=12+\left(9-1\right)10`
`a_9=12+\left(8\right)10`
`\begin{gathered} a_9=12+80 \\ a_9=92 \end{gathered}`

Therefore;

ANSWER:

`\begin{gathered} \lparen a) \\ Multiply\text{ the previous term by 2} \\ \lparen b) \\ Add\text{ 10 to the previous term} \\ \lparen\left(c\right) \\ a_n=ar^(n-1) \\ \left(d\right) \\ a_n=a+\left(n-1\right)d \\ \left(e\right) \\ B9\text{ is greater than }A9 \end{gathered}`