Answer:
[tex]a_n=78-6n[/tex]
[tex]a_{14}=78-6(14)[/tex]
Step-by-step explanation:
To solve this, we are using the formula for the nth term of an arithmetic progression:
[tex]a_n=a_1+(n-1)d[/tex]
where
[tex]a_1[/tex] is the first term of the progression
[tex]d[/tex] is the difference
[tex]n[/tex] is position of the term in the progression
We know for our problem that the bottom row contains 72 bricks, so [tex]a_1=72[/tex]. We also know that each row above decreases by 6 bricks, so the difference is -6 ([tex]d=-6[/tex]).
Replacing the values:
[tex]a_n=72+(n-1)(-6)[/tex]
[tex]a_n=72-6n+6[/tex]
[tex]a_n=72-6n+6[/tex]
[tex]a_n=78-6n[/tex]
Where [tex]n[/tex] is the row
Since we want to now the number of bricks in the 14th row, [tex]n=14[/tex]:
[tex]a_{14}=78-6(14)[/tex]
[tex]a_{14}=78-84[/tex]
[tex]a_{14}=-6[/tex]
Since bricks can't be negative, we can conclude that this is an impossible real-life situation.
