Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant.
  • The $n$-th term: $a_n = a_1 + (n-1)d$
  • The sum of the first $n$ terms: $S_n = \frac{n}{2}[2a_1 + (n-1)d]$

Step-by-Step Solution

Step 1: Utilize the given ratio of sums and the A.P. sum formula. We are given that for an A.P., $\frac{S_p}{S_q} = \frac{p^2}{q^2}$, where $p \ne q$. We substitute the formula for $S_n$ into this equation: $\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^2}{q^2}$

• Why this step? This step translates the given information about the ratio of sums into an equation involving the first term ($a_1$) and the common difference ($d$), which are the fundamental parameters of an A.P.

Step 2: Simplify the equation to find a relationship between terms and indices. Cancel the $\frac{1}{2}$ from the numerator and denominator on the left side: $\frac{p[2a_1 + (p-1)d]}{q[2a_1 + (q-1)d]} = \frac{p^2}{q^2}$ Since $p \ne q$, we can divide both sides by $\frac{p}{q}$: $\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p^2}{q^2} \cdot \frac{q}{p}$ $\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$ Now, we rewrite the numerator and denominator to resemble the form of the $n$-th term formula, $a_n = a_1 + (n-1)d$. We can multiply the numerator and denominator of the left side by 2: $\frac{2(2a_1 + (p-1)d)}{2(2a_1 + (q-1)d)} = \frac{p}{q}$ $\frac{4a_1 + 2(p-1)d}{4a_1 + 2(q-1)d} = \frac{p}{q}$ This can be rewritten as: $\frac{2a_1 + (2(p-1))d/2}{2a_1 + (2(q-1))d/2} = \frac{p}{q}$ This form isn't directly helpful for relating to $a_n$. Instead, let's manipulate the expression $2a_1 + (k-1)d$ to match $a_n$. We have $\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$. Let's rewrite the numerator as $2 \left( a_1 + \frac{p-1}{2}d \right)$ and the denominator as $2 \left( a_1 + \frac{q-1}{2}d \right)$. $\frac{2 \left( a_1 + \frac{p-1}{2}d \right)}{2 \left( a_1 + \frac{q-1}{2}d \right)} = \frac{p}{q}$ $\frac{a_1 + \left(\frac{p-1}{2}\right)d}{a_1 + \left(\frac{q-1}{2}\right)d} = \frac{p}{q}$ For this to hold true, the term in the numerator must be proportional to $p$ and the term in the denominator proportional to $q$. Specifically, if we want to relate this to the ratio of terms $a_k$ and $a_m$, we can see a pattern. If we set $a_k = a_1 + (k-1)d$ and $a_m = a_1 + (m-1)d$, then by comparing the structure: $k-1 = \frac{p-1}{2} \quad \text{and} \quad m-1 = \frac{q-1}{2}$ Solving for $p$ and $q$: $2(k-1) = p-1 \implies p = 2(k-1) + 1 = 2k - 2 + 1 = 2k-1$ $2(m-1) = q-1 \implies q = 2(m-1) + 1 = 2m - 2 + 1 = 2m-1$ Thus, the ratio of the $k$-th term to the $m$-th term is given by: $\frac{a_k}{a_m} = \frac{p}{q} = \frac{2k-1}{2m-1}$

• Why this step? This crucial step establishes a general formula relating the ratio of any two terms ($a_k, a_m$) to their indices ($k, m$), given the specific condition on the sums. This makes finding the ratio of specific terms straightforward.

Step 3: Calculate the ratio $\frac{a_6}{a_{21}}$. We need to find $\frac{a_6}{a_{21}}$. Using the derived formula $\frac{a_k}{a_m} = \frac{2k-1}{2m-1}$, we set $k=6$ and $m=21$. $\frac{a_6}{a_{21}} = \frac{2(6)-1}{2(21)-1}$ $\frac{a_6}{a_{21}} = \frac{12-1}{42-1}$ $\frac{a_6}{a_{21}} = \frac{11}{41}$

• Why this step? This step applies the general relationship derived in Step 2 to the specific terms required by the question.

Common Mistakes & Tips

• Incorrectly relating sums to terms: A common error is to assume a direct proportionality between the sums and terms (e.g., $\frac{S_p}{S_q} = \frac{p}{q}$ implies $\frac{a_p}{a_q} = \frac{p}{q}$). Always use the sum formula and simplify carefully.
• Algebraic errors: Mistakes in simplifying fractions or solving for $p$ and $q$ in terms of $k$ and $m$ can lead to incorrect answers.
• Shortcut understanding: The derived relationship $\frac{a_k}{a_m} = \frac{2k-1}{2m-1}$ is a powerful shortcut for problems of this type. Understanding why it works (as shown in Step 2) is more important than just memorizing it.

Summary

The problem provides a specific ratio for the sums of terms in an arithmetic progression: $\frac{S_p}{S_q} = \frac{p^2}{q^2}$. By substituting the formula for the sum of an A.P. and performing algebraic simplification, we derived a general relationship between the ratio of any two terms $a_k$ and $a_m$, and their indices: $\frac{a_k}{a_m} = \frac{2k-1}{2m-1}$. Applying this formula to find the ratio of the 6th term to the 21st term, i.e., $\frac{a_6}{a_{21}}$, yields $\frac{2(6)-1}{2(21)-1} = \frac{11}{41}$.

The final answer is $\boxed{\frac{11}{41}}$ which corresponds to option (D).