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JEE Mains · Maths · STD 11 - 8. sequence and series

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

  1. A \(\frac{{41}}{{11}}\)
  2. B \(\frac{{31}}{{121}}\)
  3. C \(\frac{{11}}{{41}}\)
  4. D \(\frac{{121}}{{1861}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{{31}}{{121}}\)

Step-by-step Solution

Detailed explanation

\(\frac{{{a_1} + {a_2} + {a_3} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + .... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}}\) \( \Rightarrow \frac{{{a_1} + {a_2}}}{{{a_1}}} = \frac{8}{1}\,\, \Rightarrow {a_1} + \left( {{a_1} + d} \right) = 8{a_{{\kern 1pt} 1}}\)…
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