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