Question:

Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression (AP) with first term $a_1$ and common difference $d$. The sum of the first $n$ terms of an AP is given by the formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. The given condition is: $$ \frac{a_1 + a_2 + \ldots + a_p}{a_1 + a_2 + \ldots + a_q} = \frac{p}{q} $$ This can be written as $\frac{S_p}{S_q} = \frac{p}{q}$. Substituting the sum formula: $$ \frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p}{q} $$ $$ \frac{p(2a_1 + (p-1)d)}{q(2a_1 + (q-1)d)} = \frac{p}{q} $$ For the subsequent derivation to hold, we consider the ratio of the average of the first $p$ terms to the average of the first $q$ terms, which is $\frac{S_p/p}{S_q/q} = \frac{p}{q}$. This simplifies the equation to: $$ \frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q} $$ Cross-multiplying: $$ q[2a_1 + (p-1)d] = p[2a_1 + (q-1)d] $$ $$ 2qa_1 + q(p-1)d = 2pa_1 + p(q-1)d $$ Rearranging terms to group $a_1$ and $d$: $$ 2qa_1 - 2pa_1 = p(q-1)d - q(p-1)d $$ $$ 2a_1(q-p) = d[pq - p - (pq - q)] $$ $$ 2a_1(q-p) = d[pq - p - pq + q] $$ $$ 2a_1(q-p) = d(q-p) $$ Since $p \neq q$, we know that $q-p \neq 0$. Thus, we can divide both sides by $(q-p)$: $$ 2a_1 = d $$ Now, we express the $n$-th term of the AP, $a_n$, using the relationship $d=2a_1$: $$ a_n = a_1 + (n-1)d $$ Substitute $d=2a_1$: $$ a_n = a_1 + (n-1)(2a_1) $$ Factor out $a_1$: $$ a_n = a_1[1 + 2(n-1)] $$ $$ a_n = a_1(1 + 2n - 2) $$ $$ a_n = a_1(2n-1) $$ Next, we calculate the terms $a_6$ and $a_{21}$: For $n=6$: $$ a_6 = a_1(2 \cdot 6 - 1) = a_1(12 - 1) = 11a_1 $$ For $n=21$: $$ a_{21} = a_1(2 \cdot 21 - 1) = a_1(42 - 1) = 41a_1 $$ The ratio $\frac{a_6}{a_{21}}$ is: $$ \frac{a_6}{a_{21}} = \frac{11a_1}{41a_1} = \frac{11}{41} $$ The required value is the square of this ratio: $$ \left(\frac{a_6}{a_{21}}\right)^2 = \left(\frac{11}{41}\right)^2 = \frac{121}{1681} $$