Question

Consider the sequence 5, 9, 13, 17, 21, ... ..(a) Find the 10th term of the sequence.(b) Find, in terms of n, a formula for the nth term of the sequence.(c) What is the smallest term of the sequence that is greater than 100?(d) Work out the difference between the 11th term and 17th term in the sequence.

Answer 1

(a) To find the 10th term, we could start by writing the formula, but since this is item (b) we will simply identify the pattern and calculate by hand.

From the sequence, we can see that the difference between sequential term is constant:

`\begin{gathered} 9-5=4 \\ 13-9=4 \\ 17-13=4 \\ 21-17=4 \\ \ldots \end{gathered}`

So, to find the 10th term, we can continue to add 4 until we get there.

21 is the 5th term,

`\begin{gathered} 6th\colon21+4=25 \\ 7th\colon25+4=29 \\ 8th\colon29+4=33 \\ 9th\colon33+4=37 \\ 10th\colon37+4=41 \end{gathered}`

So, the 10th term is 41.

(b) Now, to work the formula out, we need to start from the first term, 5, and add 4 a number of times so that we get to the term we want, n. If we add 4 once, we will get to the second term, if we add twice, we will get to the third term.

So, if we add n - 1 times, we will get to the nth term.

Thus, the formula is the first term plus 4 times n - 1:

`a(n)=5+4(n-1)`

We can see that the 10th term is indeed 41 using the formula:

`a(10)=5+4(10-1)=5+4\cdot9=5+36=41`

So, the formula is:

`a(n)=5+4(n-1)`

(c) The formula only works for integer n. If we pick a number to put in place of a but this number is not on the sequence, we will get a not integer n. although this is not defined, we can use this not integer n to find the closest term.

So, if we want a term close to 100, we can set a(n) = 100 and find n:

`\begin{gathered} 100=5+4(n-1) \\ 4(n-1)=100-5 \\ 4(n-1)=95 \\ n-1=(95)/(4) \\ n=(95)/(4)+1 \\ n=(95)/(4)+(4)/(4) \\ n=(99)/(4) \end{gathered}`

So, since 100 is not on the sequence we got to a fractional n. However, the closest integer n is:

`n=(100)/(4)=25`

And this n gives the term:

`\begin{gathered} a(25)=5+4(25-1) \\ a(25)=5+4\cdot24 \\ a(25)=5+96 \\ a(25)=101 \end{gathered}`

So, the smallest term of the sequence that is greater than 100 is the 25th term and its value is 101.

(d) We can use the formula to answer this.

we have:

`\begin{gathered} a(17)=5+4(17-1)=5+4\cdot16=5+64=69 \\ a(11)=5+4(11-1)=5+4\cdot1=5+40=45 \\ a(17)-a(11)=69-45=24 \end{gathered}`

So, the difference between the 11th and the 17th term is 24.