Question

The first four terms of a sequence are shown below.

9, 5, 1, -3

Which of the following functions best defines this sequence?

f(1) = 9, f(n + 1) = f(n) - 4; for n ≥ 1

f(1) = 9, f(n + 1) = f(n) + 4; for n ≥ 1

f(1) = 9, f(n + 1) = f(n) - 5; for n ≥ 1

f(1) = 9, f(n + 1) = f(n) + 5; for n ≥ 1

PLEASE HELP ME

Answer 1

HELP ME PLEASE ASAP

Answer 2

You have to work it out. \(f(1) = 9\). \(f(2) = 5\). Clearly the expression for \(f(n+1)\) has to involve subtracting something from \(f(n)\), no?

Answer 3

f(1) = 9, f(n + 1) = f(n) - 5; for n ≥ 1

Answer 4

Well, is 9-5=5?

Answer 5

\[f(1) = 9\]\[f(2) = 5\]\[f(n+1) = f(n)-5\]\[f(2) = f(1) - 5\]\[5=9-5\]\[5=4\]Uh, oops.

Answer 6

If you look at the numbers in the sequence,
9-5 = 4
5-1 = 4
1-(-3) = 4

do you see a pattern developing here? A sequence where the next term is produced by adding or subtracting a constant to/from the previous term is called an arithmetic sequence. The constant is called the common difference. You can write the formula to give you any term like this:

\[a_n = a_1 + (n-1)d\]where \(a_1\) (which equals \(f(1)\) in this problem) is the first term, \(n\) is the number of the term you wish to find, \(a_n\) is the value of the term you wish to find, and \(d\) is the common difference.

If we were trying to write this sequence in that form, we would do as follows:

\[a_1 = 9\]\[a_2 = 5\]\[a_2 = a_1 + (2-1)d\]\[5 = 9 + (2-1)d\]\[5 = 9 + 1d\]\[5-9 = 9-9+1d\]\[-4=1d\]\[d = -4\]So our formula would be \[a_n = 9 + (n-1)(-4)\]or\[a_n = 9-4(n-1)\]or if you wanted it even simpler, at the cost of hiding the origin, you could write it as\[a_n = 13-4n\]

Here it is in action:
\[a_1 = 9-4(1-1) = 9 - 0 = 9\]\[a_2 = 9-4(2-1) = 9-4(1) =9-4 = 5\]\[a_3 = 9-4(3-1) = 9-4(2) =9-8= 1\]\[a_4 = 9-4(4-1) = 9-4(3) = 9-12 = -3\]\[a_5 = 9-4(5-1) = 9-4(4) = 9-16=-7\]