To find the coefficients of the 5th, 6th, and 7th terms in the expansion of (1 + x)^n, where n is a natural number, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be expressed as:

(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + C(n, 2)a^(n-2) b^2 + ... + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n

Where C(n, k) denotes the binomial coefficient, which is the number of ways to choose k items from a set of n distinct items.

Now, let's apply the binomial theorem to expand (1 + x)^n:

(1 + x)^n = C(n, 0)1^n x^0 + C(n, 1)1^(n-1) x^1 + C(n, 2)1^(n-2) x^2 + ... + C(n, n-1)1^1 x^(n-1) + C(n, n)1^0 x^n

Simplifying this expression, we get:

(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n-1)x^(n-1) + C(n, n)x^n

Now, we need to find the coefficients of the 5th, 6th, and 7th terms. Let's denote these coefficients as a, b, and c respectively.

The 5th term is C(n, 4)x^4, the 6th term is C(n, 5)x^5, and the 7th term is C(n, 6)x^6.

Since the coefficients of these terms are in arithmetic progression, we can write the following equations:

b - a = c - b (1)
c - b = d - c (2)

Substituting the expressions for the coefficients, we get:

C(n, 5)x^5 - C(n, 4)x^4 = C(n, 6)x^6 - C(n, 5)x^5 (3)
C(n, 6)x^6 - C(n, 5)x^5 = C(n, 7)x^7 - C(n, 6)x^6 (4)

Now, let's solve these equations to find the value of n.

Cancelling out the common terms and rearranging the equations, we get:

C(n, 6)x^2 = C(n, 5)x (5)
C(n, 7)x^2 = C(n, 6) (6)

Dividing equation (5) by equation (6), we have:

C(n, 6)x^2 / C(n, 7)x^2 = C(n, 5)x / C(n, 6) (7)

Simplifying this equation, we get:

(n - 5) / (n - 6) = x (8)

Since x can take any value