Question: The pth term of an Ap is (3p-1)/6 the sum of the first n terms of sn a. P is? Explain in details.

Solution:

Definition: Arithmetic Progression (AP) is a series in which each term is the sum of the preceding term(s) and a constant, called the common difference.

Formula: The nth term of an AP is given by the formula: a + (n-1)d, where a is the first term and d is the common difference.

Given that the pth term of an AP is (3p-1)/6, we can write:

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

Multiplying both sides by 6, we get:

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

Simplifying the above equation, we get:

3p - 6d = 6a - 1

Now, let's find the sum of the first n terms of the AP, denoted by Sn.

Formula: The sum of the first n terms of an AP is given by the formula: Sn = (n/2)(2a + (n-1)d)

Substituting the value of a (which we got from the equation above), we get:

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

Simplifying the above equation, we get:

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

Multiplying both sides by 6, we get:

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

Simplifying the above equation, we get:

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

Multiplying both sides by 3, we get:

18Sn = n[6p + 6(n-1)d - 2]

Simplifying the above equation, we get:

18Sn = 6np + 6n(n-1)d - 2n

Now, let's substitute the value of 6a - 1 (which we got from the equation above) in the above equation, we get:

18Sn = 18np - 18p + 3

Simplifying the above equation, we get:

6Sn = 6np - 6p + 1

Now, let's substitute the value of Sn (which we got from the equation above) in the above equation, we get:

3n[(3p-1)/2] = 6np - 6p + 1

Simplifying the above equation, we get:

9p - 3 = 12np - 12p + 2

Simplifying the above equation, we get:

12p(n-1) = 4

Therefore, p(n-1) = 1/3

Since p and n are both integers, n-1 must be a multiple of 3. Therefore, we can write:

n-1 = 3k, where k is an integer.

Substituting the above value in p(n-1) = 1/3, we get