I have a question about this problem, please help

Question

*So the example question in my lesson is

The 4th term of an arithmetic sequence is 13 and the 8th term is 21. What is the expression for this arithmetic sequence?

a8 = a4 + 4d because a8 is 4 common differences from a4.

21 = 13 + 4d

d = 2

We can also see that a1 = a4 + (-3)d, so

a1 = 13 + (-3)(2) = 7

and c = a1 - d = 7 - 2 = 5

Then, an = dn + c = 2n + 5, so the expression is an = 2n + 5

*Where did the negative three come from?? and what is the general equation to find a1?

Answer 1

The general expression for an arithmetic sequence is a_n = a₁ + (n-1)d

For this problem: a₄ = 13 = a₁ + 3d and

a₈ = 21 = a₁ + 7d

So you have 2 linear equations in 2 unknowns.

Subtract a₄ from a₈ to get 8 = 4d => d = 2 just as you said.

Then 13 = a₁ +3*2 => a₁ = 7 just as you said.

The sequence is a_n = 7 + 2*(n-1).

Now where did you get crossed up and what is your c?