Given: The Pth term of an AP is (3P-1)/6.

To find: The sum of the first n term of the AP.

Solution:

Step 1: Find the common difference (d) of the AP.

Let the Pth term of the AP be aP, then we have:

aP = (3P-1)/6

aP-1 = (3(P-1)-1)/6 = (3P-4)/6

Using the formula for the nth term of an AP, we have:

aP = a1 + (P-1)d

aP-1 = a1 + (P-2)d

Substituting the values of aP, aP-1, and simplifying, we get:

d = (aP - aP-1) = 1/2

Therefore, the common difference of the AP is 1/2.

Step 2: Find the sum of the first n terms of the AP.

Using the formula for the sum of the first n terms of an AP, we have:

Sn = n/2[2a1 + (n-1)d]

Substituting the value of d and simplifying, we get:

Sn = n/2[a1 + a1 + (n-1)d]

Sn = n/2[2a1 + (n-1)(1/2)]

Sn = n/2[2a1 + n/2 - 1/2]

Sn = n/4[4a1 + n - 1]

Substituting the value of a1, we get:

Sn = n/4[4(1/2) + n - 1]

Sn = n/4[n/2 + 1]

Sn = n(n+2)/8

Therefore, the sum of the first n terms of the AP is n(n+2)/8.

Answer: The sum of the first n terms of the AP is n(n+2)/8.