65/9 = p + 1/9
Subtract 1/9 from both sides
64/9 = p
We have an arithmetic progression:
An arithmetic progression is a sequence of numbers such that the diference between the consecutive terms is constant.
Explicit formula:
an=a₁+(n-1)*d
a₁ is the first term
d is de common diference
n is the numbers of terms
In this case:
a₁=-9
d=an-an-1=a₂-a₁=-16-(-9)=-16+9=-7
an=-9+(n-1)*(-7)
an=-9-7n+7
an=-7n+-2
a₁=-7*1-2=-9
a₂=-7*2-2=-14-2=-16
a₃=-7*3-2=-21-2=-23
a₄=-7*4-2=-28-2=-30
a₅=-7*5-2=-35-2=-37
a₆=-7*6-2=-42-2=-44
Answer: -4
Step-by-step explanation:
f(x) = -x+2
-6+2
=-4