Given that (n+1), (n-1), and 4 are terms in an arithmetic sequence.

Let (r) be the common difference between the terms.

Then, we know that:

==> (n-1) = (n+1) + r

==> n-1 = n+1 + r

==> -1= 1 + r

==> **r= -2**..............(1).

Also, we know that:

==> 4 = (n-1) + r

But r = -2.

==> 4 = (n-1) + -2

==> 4 = n-1 -2

==> 4 = n -3

==> **n = 7.**

==> **(n+1) , (n-1) , 4 = 8, 6, 4 are terms of an arithmetic sequence where r = -2 is the common difference.**

The terms (n+1), (n-1), 4 are the consecutive terms of a arithmetical progression (sequence) if and only if the middle term is the arithmetical mean of the neighbor terms:

n - 1 = [(n+1) + 4]/2

We'll multiply by 2 both sides:

2n - 2 = n + 1 + 4

We'll combine like terms from the right side:

2n - 2 = n + 5

We'll subtract n and add 2 both sides:

**n = 7**

**The terms of the arithmetical sequence, whose common difference is d = -2 are: 7+1 = 8 , 7-1 = 6 , 4.**