Given the terms:

(x-2) , 6, (2x-3) are terms of an arithmetical progression.

Then, we will assume that "r" is the common difference between terms.

==> 6 = (x-2) + r

==> x+r = 8.............(1)

==> (2x-3) = 6 + r

==> 2x -r = 9 ..............(2)

Now we will solve the system using the elimination method.

We will add (1) and (2).

==> 3x = 17

**==> x = 17/3**

==> r = 6- 11/3 = 7/3

==> (x-2) = 17/3 - 6/3 = 11/3

==> 6 = 6

==> 2x-3 = 34/3 - 9/3 = 25/3

**==> 11/3, 6, 25/3 are terms of A.P and the common difference is r = 7/3**

We have (x-2), 6, (2x-3) as consecutive terms of an AP. The consecutive terms of an AP have a common difference.

So 2x - 3 - 6 = 6 - (x - 2)

=> 2x - 9 = 6 - x + 2

=> 2x - 9 = -x + 8

=> 3x = 17

=> x = 17/3

**Therefore x is equal to 17/3.**

In an arithmetic progression( A.P), the successive terms are increasing or decreasing by the same amount called the common difference d.

=> Second term - first term = third term - second term.

The given terms of A.P is (x-2), 6 and (2x-3).

6 - (x-2) = 2x-3 - 6.

6-x+2 = 2x-9

6+2+9 = 2x+x

17 = 3x.

17/3 = x.

So x= 17/3

Therefore the first term of the A.P are x-2 = 17/3 -2 = 11/6.

The third term = 2x-3 = 2(17/3)-3 = 25/3.

Therefore the terms of the A.P are 11/3, 6 and 25/3.