The first figure has 10 small blocks
The second figure has 14 small blocks
Then the difference between the two figures is 14 - 10 = 4
Then it is an arithmatic sequance
a)
Then the next two figures will have (14 + 4 = 18 small blocks) and (18 + 4 = 22 small blocks)
You can draw them 18 = 9 x 2 and 22 = 11 x 2 small blocks
b)
The input will be the number of the figure (n)
The output is the number of the small blocks in each figure
Input: Output
1 10
2. 14
3. 18
4. 22
c)
The recursive formula of the arithmetic sequence is
[tex]a1=1^{st}term,a_n=a_{n-1}+d[/tex]
a1 is the first term
an is any term in the sequence
d is the difference between each 2 consecutive terms
Since a1 = 10
Since d = 4
Then the recursive formula is
[tex]a_1=10,a_n=a_{n-1}+4[/tex]
The explicit formula of the arithmetic sequence is
[tex]a_n=a_1+(n-1)d[/tex]
Since a1 = 10
Since d = 4
Then the explicit formula is
[tex]a_n=10+(n-1)(4)[/tex]
We will simplify it by multiplying the bracket by 4, then add the like terms
[tex]\begin{gathered} a_n=10+(n)(4)-(1)(n) \\ a_n=10+4n-4 \\ a_n=(10-4)+4n \\ a_n=6+4n \end{gathered}[/tex]
d)
You can find any real situation that something is increasing by 4 each time
Example:
The length of the plant is increasing by 4 cm each day
e)
day: length of the plant
1. 10
2. 14
3. 18
4. 22
f)
To find the number of blocks in figure 31
Substitute n by 31, then find an in the explicit formula
[tex]\begin{gathered} n=31 \\ a_{31}=6+4(31) \\ a_{31}=6+124 \\ a_{31}=130 \end{gathered}[/tex]
Figure 31 has 130 blocks
g)
To find the number of the figure that has 50 blocks
Substitute an by 50 and find n
[tex]\begin{gathered} a_n=50 \\ 50=6+4n \end{gathered}[/tex]
Subtract 6 from both sides
[tex]\begin{gathered} 50-6=6-6+4n \\ 44=4n \end{gathered}[/tex]
Divide both sides by 4 to find n
[tex]\begin{gathered} \frac{44}{4}=\frac{4n}{4} \\ 11=n \end{gathered}[/tex]
The figure that has 50 blocks is the 11 figure
