Given that 2x, 17, 3x-1 are terms is A.P

Let us assume that the common difference is r.

Then we know that:

17 = 2x + r ...........(1)

Also we know that:

3x-1 = 17 + r

=> -18 = -3x + r...........(2)

Let us solve the system.

We will subtract (2) from (1).

==> 35 = 5x

==> x = 35/5 = 7

Then x= 7

Now we will substitute and determine the terms.

==> 14 , 17, 20

**Then the terms are : 14, 17, 20 and the value of x = 7**

The terms 2x , 17 and 3x - 1 are in an arithmetical progression.

We know that for 3 consecutive terms of an AP, the second term is the arithmetic mean of the 1st and 3rd terms.

=> 17 = (2x + 3x - 1)/2

=> 17 = (5x - 1)/2

=> 5x - 1 = 34

=> 5x = 35

=> x = 35/5

=> x = 7

**The value of x = 7.**

It is given that 2x , 17 and 3x-1 are terms of an arithmetic progression. The nth term of an arithmetic progression is Tn = T + (n-1)*d where T is the first term and d is the common difference.

The first term is 2x and 17 is the second term.

17 = 2x + d...(1)

3x - 1 is the third term.

3x - 1 = 2x + 2*d

=> x = 2d + 1

Substitute this in (1), 17 = 4d + 2 + 2d = 6d + 2

d = 15/6

Substituting d = 15/6 in x = 2d+1 gives x = 7

The terms of the arithmetic progression are 14, 17 and 20