If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is:

We consider an arithmetic progression (A.P.) whose first term is denoted by $$a$$ and common difference by $$d$$. For any A.P., the formula for the $$n^{\text{th}}$$ term is

$$T_n = a + (n-1)d.$$

We are told that the nineteenth term is zero. Putting $$n = 19$$ in the formula, we obtain

$$T_{19} = a + (19-1)d = a + 18d = 0.$$

Solving this linear equation for $$a$$ gives

$$a = -18d.$$

Now we calculate the forty-ninth term. Substituting $$n = 49$$ in the general term formula, we get

$$T_{49} = a + (49-1)d = a + 48d.$$

Replacing $$a$$ by $$-18d$$, we find

$$T_{49} = -18d + 48d = 30d.$$

Next, we compute the twenty-ninth term. Putting $$n = 29$$ gives

$$T_{29} = a + (29-1)d = a + 28d.$$

Again substituting $$a = -18d$$, we obtain

$$T_{29} = -18d + 28d = 10d.$$

We are asked to find the ratio of the forty-ninth term to the twenty-ninth term. Using the values just found, we have

$$\frac{T_{49}}{T_{29}} = \frac{30d}{10d}.$$

The factor $$d$$ cancels, and the numerical ratio simplifies to

$$\frac{30}{10} = 3.$$

Therefore,

$$(49\text{th term}) : (29\text{th term}) = 3 : 1.$$

Hence, the correct answer is Option C.