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8. Sequences and Series
hard
Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$ If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals
A
$\frac{{41}}{{11}}$
B
$\frac{7}{2}$
C
$\frac{2}{7}$
D
$\frac{{11}}{{41}}$
(AIEEE-2006)
Solution
$\frac{\frac{p}{2}\left|2 a_{1}+(p-1) d\right|}{\frac{q}{2}\left|2 a_{1}+(q-1) d\right|}=\frac{p^{2}}{q^{2}}$
$\Rightarrow \frac{2 a_{1}+(p-1) d}{2 a_{1}+(p-1) d}=\frac{p}{q}$
$\frac{a_{1}+\left(\frac{p-1}{2}\right) d}{a_{1}+\left(\frac{q-1}{2}\right) d}=\frac{p}{q}$
For $\frac{a_{6}}{a_{21}}, p=11, q=41$
$\Rightarrow \frac{a_{6}}{a_{21}}=\frac{11}{41}$
Standard 11
Mathematics
