Let $a_1,a_2,a_3....$ be an arithmetic progression. Question is

Let $a_1,a_2,a_3....$ be an arithmetic progression. If $\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2},p\not=q$, then find $\frac{a_6}{a_{21}}$

Answer: $\frac{11}{41}$
All options: $\frac{7}{2}$, $\frac{2}{7}$, $\frac{11}{41}$, $\frac{41}{11}$


My approach to the question is
Let $S_r$ be sum of first $r$ terms of this AP
Method 1
we know
$\frac{S_p}{S_q}=\frac{p^2}{q^2}$
So,
$\frac{\frac{p}{2}(a_1+a_p)}{\frac{q}{2}(a_1+a_q)}=\frac{p^2}{q^2}$
$\frac{a_1+a_p}{a_1+a_q}=\frac{p}{q}$
After that I am stuck.
Method 2
But after taking another look at the question, I observed
$\frac{S_6}{S_{21}}=\frac{6^2}{21^2}=\frac{36}{441}$
Also,
$\frac{S_5}{S_{20}}=\frac{5^2}{20^2}=\frac{25}{400}$
And we know $S_r-S_{r-1}=a_r$
Then subtracting numerators and denominators of both
$\frac{S_6-S_5}{S_{21}-S_{20}}=\frac{a_6}{a_{21}}=\frac{36-25}{441-400}=\frac{11}{41}$
which we can't always do in maths, I cannot understand why Method 2 gave correct answer.

$$\dfrac{p^2}{q^2}=\dfrac{p(2a_1+(p-1)d}{q(2a_1+(q-1)d}$$
$$\implies \dfrac pq=\dfrac{a_1+d(p-1)/2}{a_1+d(q-1)/2}$$
Set $\dfrac{p-1}2=6-1$
and $\dfrac{q-1}2=21-1$
