Proving $a_n = a_i + (n - i)d$ by induction

The formula for the $n$th term of an arithmetic sequence is $$a_n = a_1 + (n - 1)d \tag{1}\label{1}.$$ However, I want to prove that this can be generalized to $$a_n =a_i + (n - i)d. \label{2}\tag{2}$$

Using \eqref{1}, I can get \eqref{2} directly: \begin{align*} a_n &= a_1 + (n - 1)d \\ a_n &= a_1 + (n - i + i - 1)d \\ a_n &= a_1 + (i - 1)d + (n - i)d \\ a_n &= a_i + (n - i)d \end{align*}

How can we prove this by induction? I know that direct proofs are preferable over indirect proofs, but since this is a well-behaved sequence, a proof by induction should be achievable.

Here would be my technique: let $i$ be an arbitrary positive integer, and $a$ be an arithmetic sequence satisfying $a_{k+1} = a_k + d$ for all positive integers $k.$ Now we can prove $a_n = a_i + (n-i)d$ using induction on $n.$

Base case: if $n = i,$ then trivially our formula holds: $a_i = a_i + (i-i)d.$

Now supposing that we have $a_{n_0} = a_i + (n_0-i)d$ for some positive integer $n_0,$ then by substituting $n_0$ for $k$ in the definition of our sequence we have $a_{n_0 + 1} = a_{n_0} + d,$ so we must have $a_{n_0 + 1} = a_i + (n_0 - i)d + d = a_i + (n_0 + 1 - d),$ satisfying our formula. By induction this proves that our formula holds for integers $n > i.$

Now notice that similarly we can substitute $k = n_0 - 1,$ supposing that it is also positive, to get that $a_{n_0} = a_{n_0 - 1} + d \Rightarrow a_{n_0 - 1} = a_{n_0} - d.$ Now we must have that $a_{n_0 - 1} = a_i + (n_0 - i)d - d = a_i + (n_0 - 1 - i)d,$ also satisfying our formula. This shows that our formula holds for all integers $0 < n < i,$ and together with our other argument we've proven that our formula holds for all integers $n > 0.$

Now, because our value for $i$ is arbitrary, this argument works for all positive integers $i,$ so our formula holds for any pair of positive integers $i$ and $n.$

This is logically equivalent to induction.

Assume (2) is true. then re-write $a_n$ in terms of $a_1$ , $n$, $i$, and $d$.

Can you get back (1). If yes, the assumption is correct.

Base case: $a_1=a_1 + 0d$ is true.

Assume $P(n,i)$ is true. Then we show that $P(n+1,i)$ and $P(n,i+1)$ are both true.

$P(n+1,i)$ is the statement that $a_{n+1} = a_i+nd$ and $P(n,i+1)$ is the statement that $a_n = a_{i+1}+(n-i-1)d$.

The first is easy - add $d$ to both sides.

The second, add $d-d=0$ to both sides: $a_i +d = a_{i+1}$ and $(n-i)d-d=(n-i-1)d$.