The consecutive terms of an AP have a common difference.

So 2x + 1 - 7 = 7 - (x - 2)

=> 2x - 6 = 9 - x

=> 3x = 15

=> x = 15/3

**=> x = 5**

Given that:

(x-2) , 7 , (2x+1) are terms of an arithmetical progression.

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

Then we know that:

7 = (x-2) + r

==> 7 - r = x-2

==> x = 9 - r ................(1)

Also, we know that:

2x+1 = 7 + r

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

Now we will add (1) and (2).

==> 3x = 15

==> x = 15/3 = 5

To check:

5 -2 = 3

2x +1 = 2(5) + 1 = 11

==> 3, 7, 11 are terms of an A.P where the common difference is r = 4

**==> x = 5**