Solution:

The given expression is (3 + 7x)²⁹.

To find the coefficient of the rth and (r + 1)th terms, we use the formula:

Coefficient of the rth term = nCr (a)^(n-r) (b)^r where a = 3 and b = 7x

Coefficient of the (r + 1)th term = nC(r+1) (a)^(n-r-1) (b)^(r+1) where a = 3 and b = 7x

We know that the coefficients of the rth and (r + 1)th terms are equal. Therefore,

nCr (a)^(n-r) (b)^r = nC(r+1) (a)^(n-r-1) (b)^(r+1)

Simplifying this expression, we get:

[(n-r)/(r+1)] * [(7x)/(3)] = r+1

Multiplying both sides by (r+1)/[(n-r)/(r+1)]

7x/(n-r) = (r+1)²/(n-r)

Multiplying both sides by (n-r), we get:

7x = (r+1)²

Taking the square root of both sides, we get:

√(7x) = r+1

Therefore, r = √(7x) - 1.

To find the value of r, we need to substitute the given coefficients of the terms in the expression for r. The coefficients of the rth and (r + 1)th terms are equal. Therefore,

nCr (a)^(n-r) (b)^r = nC(r+1) (a)^(n-r-1) (b)^(r+1)

Using this expression, we can find the value of r as follows:

nCr (3)^(29-r) (7x)^r = nC(r+1) (3)^(28-r) (7x)^(r+1)

Dividing both sides by nCr (3)^(28-r) (7x)^r, we get:

3/(29-r) = (r+1)/(7x)

Multiplying both sides by (29-r)/(r+1), we get:

3(r+1) = (29-r)/(7x)

Multiplying both sides by 7x, we get:

21x(r+1) = 29-r

Expanding the expression on the left-hand side, we get:

21xr + 21x = 29 - r

Adding r to both sides, we get:

21xr + 21x + r = 29

Factoring out r on the left-hand side, we get:

r(21x + 1) = 29 - 21x

Dividing both sides by (21x + 1), we get:

r = (29 - 21x)/(21x + 1)

To find the value of r such that the coefficients of the rth and (r + 1)th terms are equal, we need to substitute this expression for r in the equation we derived earlier:

√(7x) = r+1

Substituting the expression for r, we get:

√(7x) = (29 -