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.
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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.
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