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Let a_1, a_2, a_3, … be an A.P, such that a_1+a_2+…+a_p a_1+a_2+a_3+…+a_q= p^3 q^3 ; p ≠ q. Then a_6 a_21 is equal to:

JEE Main 2013 — Mathematics Algebra

2013mcqeasy

Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be an A.P, such that a1+a2+…+apa1+a2+a3+…+aq=p3q3;p≠q\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+a_3+\ldots+a_q}=\frac{p^3}{q^3} ; p \neq qa1​+a2​+a3​+…+aq​a1​+a2​+…+ap​​=q3p3​;p=q. Then a6a21\frac{a_6}{a_{21}}a21​a6​​ is equal to:

Official previous-year question

Held on 9 Apr 2013 · Verified 6 Jul 2026.

Options

  1. A

    4111\frac{41}{11}1141​

  2. B

    31121\frac{31}{121}12131​

  3. C

    1141\frac{11}{41}4111​

  4. D

    1211861\frac{121}{1861}1861121​

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