A(n)=3+(n-1)(5)

Solution for A(n)=3+(n-1)(5) equation:

(A)=3+(A-1)(5)

We move all terms to the left:

(A)-(3+(A-1)(5))=0


We calculate terms in parentheses: -(3+(A-1)5), so:

3+(A-1)5

determiningTheFunctionDomain
(A-1)5+3

We multiply parentheses

5A-5+3

We add all the numbers together, and all the variables

5A-2

Back to the equation:

-(5A-2)


We get rid of parentheses

A-5A+2=0

We add all the numbers together, and all the variables

-4A+2=0

We move all terms containing A to the left, all other terms to the right

-4A=-2

A=-2/-4

A=1/2