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8. Sequences and Series
hard
Let $a_1, a_2 , a_3,.....$ be an $A.P$, such that $\frac{{{a_1} + {a_2} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + ..... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}};p \ne q$. Then $\frac{{{a_6}}}{{{a_{21}}}}$ is equal to
A
$\frac{{41}}{{11}}$
B
$\frac{{31}}{{121}}$
C
$\frac{{11}}{{41}}$
D
$\frac{{121}}{{1861}}$
(JEE MAIN-2013)
Solution
$\frac{{{a_1} + {a_2} + {a_3} + …. + {a_p}}}{{{a_1} + {a_2} + {a_3} + …. + {a_q}}} = \frac{{{p^3}}}{{{q^3}}}$
$ \Rightarrow \frac{{{a_1} + {a_2}}}{{{a_1}}} = \frac{8}{1}\,\, \Rightarrow {a_1} + \left( {{a_1} + d} \right) = 8{a_{{\kern 1pt} 1}}$
$ \Rightarrow d = 6{a_1}$
Now $\frac{{{a_6}}}{{{a_{21}}}} = \frac{{{a_1} + 5d}}{{{a_1} + 20d}}$
$ = \frac{{{a_1} + 5 \times 6{a_1}}}{{{a_1} + 20 \times 6{a_1}}} = \frac{{1 + 30}}{{1 + 120}} = \frac{{31}}{{121}}$
Standard 11
Mathematics
