**Sum of an Arithmetic Progression (AP)**

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. The sum of the first n terms of an AP can be calculated using the following formula:

Sn = (n/2) * [2a + (n-1)d]

Where:
- Sn represents the sum of the first n terms
- a represents the first term of the AP
- d represents the common difference between consecutive terms
- n represents the number of terms in the AP

**Finding the pth term of the AP**

The pth term of the AP is given by the formula:

Tp = a + (p-1)d

Given that the pth term of the AP is expressed as (3p - 1) / 6, we can equate it to the general formula of the pth term:

(3p - 1) / 6 = a + (p-1)d

Simplifying the equation, we get:

3p - 1 = 6a + 6(p-1)d

Rearranging the terms, we have:

3p - 6(p-1)d = 6a + 1

Expanding the brackets:

3p - 6pd + 6d = 6a + 1

Combining like terms:

(3 - 6d)p = 1 - 6d + 6a

Dividing both sides by (3 - 6d):

p = (1 - 6d + 6a) / (3 - 6d)

**Calculating the sum of the first n terms**

Now that we have the formula for the pth term, we can substitute it into the formula for the sum of the first n terms:

Sn = (n/2) * [2a + (n-1)d]

Replacing a with (3p - 1) / 6, we get:

Sn = (n/2) * [2((3p - 1) / 6) + (n-1)d]

Simplifying further:

Sn = (n/2) * [(3p - 1) / 3 + (n-1)d]

Sn = (n/6) * (3p - 1 + 3(n-1)d)

Sn = (n/6) * (3p + 3(n-1)d - 1)

Sn = (n/6) * [(3 - 6d)p + 3n - 3d - 1]

Sn = (n/6) * (3p - 6dp + 3n - 3d - 1)

Sn = (n/6) * (3p + 3n - 6dp - 3d - 1)

Therefore, the sum of the first n terms of the AP is given by (n/6) * (3p + 3n - 6dp - 3d - 1).

N/2(3n+2)