Question

In an A.P., $\frac{S_p}{S_q}=\frac{p^2}{q^2},p\ne q$, then $\frac{T_6}{T_{21}}$ is equal to

• A. $\frac{7}{2}$
• B. $\frac{2}{7}$
• C. $\frac{11}{41}$
• D. $\frac{41}{11}$

Answer 1

Answer: (C)

$\frac{11}{41}$

$\frac{S_p}{S_q}=\frac{P^2}{q^2}\Rightarrow \frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p}{q}$ ............(1)
$\frac{T_6}{T_{21}}=\frac{a+5d}{a+20d}=\frac{2a+10d}{2a+40d}$
$\therefore$ p-1 = 10 , q- 1 = 40 by (1) $\therefore \frac{p}{q}=\frac{11}{41}$