# If (x+1),3x,(4x+2) are consecutive terms of an A.P, then what is the value of x?

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We are given that (x+1), 3x, (4x+2) are in A.P. As they are A.P., consecutive terms have a common difference.

So 4x + 2 - 3x = 3x - x - 1

=> 4x - 3x + 2 = 2x - 1

=> x + 2 = 2x - 1

=> 2x - x = 2 + 1

=> x = 3

**Therefore if x = 3, (x+1), 3x, (4x+2) are in A.P**

In an AP the successive terms have the same common difference.

Therefore the second term - first term = thirdrd term - second term.

So 3x-(x+1) = 4x+2-3x.

2x-1 = x+2.

2x-x = 2+1

x = 3.

Therefore the value of **x = 3.**

The given 3 terms of the A.P are x+1 = 4, 3x= 3*3 = 9 and 4x+2 = 4*3+2 = 14.

The common difference of the A.P is 5.