Let a_1, a_2 , a_3,..... be an A.P, such that | vedclass

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Question

$\frac{{{a_1} + {a_2} + {a_3} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + .... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}}$

$ \Rightarrow \frac{{{a_1} + {a_2

Vedclass · Accepted Answer

$\frac{{31}}{{121}}$

Vedclass · Answer

$\frac{{{a_1} + {a_2} + {a_3} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + .... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}}$

$ \Rightarrow \frac{{{a_1} + {a_2}}}{{{a_1}}} = \frac{8}{1}\,\, \Rightarrow {a_1} + \left( {{a_1} + d} ight) = 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 	imes 6{a_1}}}{{{a_1} + 20 	imes 6{a_1}}} = \frac{{1 + 30}}{{1 + 120}} = \frac{{31}}{{121}}$

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

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