Question

TEST Unit 3B
Aaron Medina
#21.
In a sequence of numbers, "a" Subscript, 3 , Baseline equals 0 , "a" Subscript, 4 , Baseline equals 4 , "a" Subscript, 5 , Baseline equals 8 , "a" Subscript, 6 , Baseline equals 12 , "a" Subscript, 7 , Baseline equals 16.a3=0, a4=4, a5=8, a6=12, a7=16.

Based on this information, which equation can be used to find the "n" Superscript, "t" "h" , Baselinenth term in the sequence, "a" Subscript, "n" , Baselinean?


"a" Subscript, "n" , Baseline equals negative 4 "n" minus 6
"a" Subscript, "n" , Baseline equals negative 4 "n" minus 12
"a" Subscript, "n" , Baseline equals 4 "n" minus 12
"a" Subscript, "n" , Baseline equals 4 "n" minus 8

Answer 1

To find the equation for the sequence given:

- \( a_3 = 0 \)
- \( a_4 = 4 \)
- \( a_5 = 8 \)
- \( a_6 = 12 \)
- \( a_7 = 16 \)

We can observe how the sequence progresses. The differences between the terms are consistent:

- \( a_4 - a_3 = 4 - 0 = 4 \)
- \( a_5 - a_4 = 8 - 4 = 4 \)
- \( a_6 - a_5 = 12 - 8 = 4 \)
- \( a_7 - a_6 = 16 - 12 = 4 \)

The common difference is \( 4 \), indicating that it is a linear sequence.

The general form of a linear sequence can be defined as:

\[ a_n = a_1 + (n - 1) \cdot d \]

where \( d \) is the common difference (4 in this case).

To find \( a_1 \) (the first term), we can work backwards from the known terms:

- The term \( a_3 \) is \( 0 \) (which corresponds to \( n = 3 \)):

Let's find \( a_1 \):

- From \( a_3 \):
\( a_3 = a_1 + (3 - 1) \cdot 4 = a_1 + 2 \cdot 4 = a_1 + 8 \)
Setting this equal to \( 0 \):
\[ a_1 + 8 = 0 \]
\[ a_1 = -8 \]

Now we can express the \( n \)th term as follows:

\[ a_n = -8 + (n - 1) \cdot 4 \]
\[ a_n = -8 + 4n - 4 \]
\[ a_n = 4n - 12 \]

Thus, the equation that can be used to find the \( n \)th term in the sequence is:

\[ a_n = 4n - 12 \]

Therefore, the correct answer is:

**"a" Subscript, "n" , Baseline equals 4 "n" minus 12.**