Key Concepts and Formulas

• General Term of an Arithmetic Progression (A.P.): The $n^{th}$ term ($T_n$) of an A.P. with first term $a$ and common difference $d$ is given by $T_n = a + (n-1)d$.
• Ratio of Terms: To find the ratio of two terms, say $T_m : T_n$, we compute $\frac{T_m}{T_n}$.

Step-by-Step Solution

• Step 1: Utilize the given information about the 19th term. We are given that the 19th term of the A.P. is zero. Using the formula for the $n^{th}$ term, $T_n = a + (n-1)d$, for $n=19$: $T_{19} = a + (19-1)d$ $0 = a + 18d$ This equation establishes a fundamental relationship between the first term ($a$) and the common difference ($d$).

• Step 2: Express the first term ($a$) in terms of the common difference ($d$). From the equation obtained in Step 1 ($a + 18d = 0$), we can rearrange it to solve for $a$: $a = -18d$ This will be useful for simplifying expressions for other terms.

• Step 3: Write the general expressions for the 49th and 29th terms. Using the general term formula $T_n = a + (n-1)d$: For the 49th term ($n=49$): $T_{49} = a + (49-1)d = a + 48d$ For the 29th term ($n=29$): $T_{29} = a + (29-1)d = a + 28d$

• Step 4: Substitute the expression for 'a' into the formulas for $T_{49}$ and $T_{29}$. Now, we substitute $a = -18d$ (from Step 2) into the expressions for $T_{49}$ and $T_{29}$ (from Step 3): For $T_{49}$: $T_{49} = (-18d) + 48d = 30d$ For $T_{29}$: $T_{29} = (-18d) + 28d = 10d$

• Step 5: Calculate the ratio $T_{49} : T_{29}$. We need to find the ratio $\frac{T_{49}}{T_{29}}$. $\frac{T_{49}}{T_{29}} = \frac{30d}{10d}$ The problem states that the A.P. is non-zero. Since $T_{19}=0$, if $d$ were 0, then $a$ would also have to be 0, making all terms zero, which contradicts "non-zero A.P.". Therefore, $d \neq 0$. We can cancel $d$ from the numerator and denominator: $\frac{T_{49}}{T_{29}} = \frac{30}{10} = \frac{3}{1}$ Thus, the ratio of the 49th term to the 29th term is $3:1$.

Common Mistakes & Tips

• Misinterpreting "Non-zero A.P.": This condition is vital. It ensures that the common difference $d$ cannot be zero if a term is zero, allowing us to cancel $d$ in calculations.
• Algebraic Errors: Be careful with signs and calculations when substituting and simplifying expressions involving $a$ and $d$.
• Direct Relationship of Term Numbers: Notice that $T_n = 0$ implies $a = -(n-1)d$. This leads to a general property: if $T_m = 0$, then $T_p : T_q = (p-m) : (q-m)$. In this case, $m=19$, $p=49$, $q=29$, so the ratio is $(49-19) : (29-19) = 30 : 10 = 3 : 1$.

Summary

The problem provides the value of the 19th term of an A.P. and asks for the ratio of the 49th term to the 29th term. By using the general term formula, we established a relationship between the first term ($a$) and the common difference ($d$) from the given information ($T_{19}=0$). Substituting this relationship into the formulas for $T_{49}$ and $T_{29}$ allowed us to express both terms solely in terms of $d$. The ratio was then calculated by dividing these expressions, leading to the result $3:1$.

The final answer is $\boxed{3 : 1}$ which corresponds to option (A).