Let the terms of an arithmetic progression be | vedclass

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(B) Given $\frac{S_p}{S_q} = \frac{p^2}{q^2}$.Using the formula for the sum of $n$ terms of an $AP$,$S_n = \frac{n}{2}[2a_1 + (n-1)d]$,we have:$\frac{

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$11/41$

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(B) Given $\frac{S_p}{S_q} = \frac{p^2}{q^2}$.Using the formula for the sum of $n$ terms of an $AP$,$S_n = \frac{n}{2}[2a_1 + (n-1)d]$,we have:$\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^2}{q^2}$$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$Dividing the numerator and denominator by $2$,we get:$\frac{a_1 + \frac{(p-1)}{2}d}{a_1 + \frac{(q-1)}{2}d} = \frac{p}{q} \dots (1)$We need to find $\frac{a_6}{a_{21}} = \frac{a_1 + 5d}{a_1 + 20d}$.Comparing this with equation $(1)$,we set $\frac{p-1}{2} = 5 \Rightarrow p = 11$ and $\frac{q-1}{2} = 20 \Rightarrow q = 41$.Substituting these values into equation $(1)$,we get $\frac{a_6}{a_{21}} = \frac{11}{41}$.

Let the terms of an arithmetic progression be $a_1, a_2, a_3, \dots$. If $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^2}{q^2}$,where $p \neq q$,then $\frac{a_6}{a_{21}} = \dots$

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