Answer:
f(6)=f(1)+15
Step-by-step explanation:
Ok if f(6)-f(5)=3, then f(n)-f(n-1)=3 for any integer n greater than or equal to 2.
f(6)-f(1)
=(f(6)-f(5))+(f(5)-f(4))+(f(4)-f(3))+(f(3)-f(2))+(f(2)-f(1))
=(3) + (3) +(3) +(3) +(3)
=5(3)
=15
So the answer is the third one:
f(6)=f(1)+15
Arithmetic sequences are linear.
So no matter the points we choose, we should get the same slope.
[tex]\frac{f(6)-f(5)}{6-5}=\frac{f(6)-f(1)}{6-1}=3[/tex]
Both slopes are 3 since we were given term-previous term is 3.
[tex]\frac{f(6)-f(1)}{6-1}=3[/tex]
[tex]\frac{f(6)-f(1)}{5}=3[/tex]
Multiply both sides by 5:
[tex]f(6)-f(1)=5(3)[/tex]
[tex]f(6)-f(1)=15[/tex]
[tex]f(6)=f(1)+15[/tex]
