# What is x if x+1, x-1, 4 are terms of an arithmetic series?

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We are given that x+1, x-1, 4 form an arithmetic series.

Therefore we have 4 - (x - 1) = (x - 1) - (x + 1)

=> 4 - x + 1 = x - 1 - x - 1

=> 5 - x = -2

=> x = 5 + 2

=> x = 7

**Therefore x = 7.**

Check:

7 + 1 = 8, 7 - 1 = 6 and 4 have a common difference of 2.

x+1, x-1 and 4 are the terms of an arithmetic progression.

If x+1, x-1 and 4 are the consecutive terms of an arithmetic series, then .

second term - first term = third term - second term.

=> x-1-(x+1 ) = 4- (x-1) is the common diffrence.

=> -2 = 4-x+1.

=> x= 4+1+2=7.

Therefore x= 7.

To determine the terms of the given arithmetical progression, we'll have to find out the value of x, first.

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

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

We'll multiply by 2 both sides:

2x - 2 = x + 1 + 4

We'll combine like terms from the right side:

2x - 2 = x + 5

We'll subtract x+5:

2x - 2 - x - 5 = 0

x - 7 = 0

We'll add 7 both sides:

**x = 7**

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