Explanation:

To find the coefficient of the second, third, and fourth terms in the binomial expansion of (1+x)^n, we need to understand the general formula for the binomial expansion and the concept of the arithmetic progression (A.P.).

Binomial Expansion:
The binomial expansion of (1+x)^n can be calculated using the binomial theorem. According to the binomial theorem, the expansion is given by:

(1+x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + C(n,3)x^3 + ... + C(n,n)x^n

where C(n,r) represents the binomial coefficient, given by:

C(n,r) = n! / (r!(n-r)!)

Arithmetic Progression (A.P.):
An arithmetic progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant. The common difference between the terms is denoted by 'd'.

For an A.P., the nth term (Tn) can be calculated using the formula:
Tn = a + (n-1)d

where 'a' is the first term and 'd' is the common difference.

Coeficients in A.P.:
Now, to determine the coefficients of the second, third, and fourth terms in the binomial expansion, we need to find the values of r for which the binomial coefficients form an arithmetic progression.

From the binomial expansion formula, we can see that the coefficients C(n,1), C(n,2), and C(n,3) correspond to the second, third, and fourth terms respectively.

Let's calculate the difference between the consecutive binomial coefficients:

d1 = C(n,2) - C(n,1) = (n! / (2!(n-2)!)) - (n! / (1!(n-1)!))
= (n(n-1)(n-2)! / (2(n-2)!)) - (n(n-1)(n-2)! / (1(n-1)!))
= n(n-1)(n-2)! / (2(n-2)!) - n(n-1)(n-2)! / (1(n-1)!)
= n(n-1) / 2 - n(n-1)
= -n(n-1) / 2

d2 = C(n,3) - C(n,2) = (n! / (3!(n-3)!)) - (n! / (2!(n-2)!))
= (n(n-1)(n-2)! / (3(n-3)!)) - (n(n-1)(n-2)! / (2(n-2)!))
= n(n-1) / 3 - n(n-1) / 2
= -n(n-1) / 6

We can observe that the differences d1 and d2 are both negative and have a common factor of -n(n-1). This implies that the coefficients C(n,1), C(n,2), and C(n,3) form an arithmetic progression with a common difference of -n(n-1)/2.

Conclusion:
Since the coefficients of the second,