Answer:
[tex]\large\boxed{1.\ f(1)=135}\\\boxed{2.\ f(6)=115}\\\boxed{3.\ f(26)=25}\\\boxed{4.\ f(n)=139-4n}[/tex]
Step-by-step explanation:
[tex]f(1)=135\\f(2)=135-4=131\\f(3)=131-4=127\\f(4)=127-4=123\\f(5)=123-4=119\\\vdots\\\\\text{It's an arithmetic sequence with firs term = 135 and the common}\\\text{difference d = -4.}\\\text{The formula of arithmetic sequence: }\\\\f(n)=f(1)+(n-1)d\\\\\text{We have}\ f(1)=135\ \text{and}\ =-4.\ \text{Substitute:}\\\\f(n)=135+(n-1)(-4)=135+(n)(-4)+(-1)(-4)\\=135-4n+4=139-4n\\\\\boxed{f(n)=139-4n}[/tex]
[tex]\text{Put n = 6, n=26 to the formula:}\\\\f(6)=139-4(6)=139-24=115\\\\f(26)=139-4(26)=139-104=25[/tex]
