Verified Solution • 2025–26 Session moderate

QJEEM06Ma42aa: JEE Main 2006 Mathematics - Class 11 JEEM06Ma42aa Solution: Let , , .....be terms on A.P. If equals

Let a1{a_1} , a2{a_2} , a3{a_3} .....be terms on A.P. If a1+a2+.....apa1+a2+.....aq=p2q2, p≠q, then a6a21 {{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p \ne q,\,then\,{{{a_6}} \over {{a_{21}}}}\, equals

Multiple Choice Options

  • A 4111{{41} \over {11}}
  • B 72{7 \over 2}
  • C 27{2 \over 7}
  • D 1141{{11} \over {41}}

Step-by-Step Verified Solution

p2[2a1+(p−1)d]q2[2a1+(q−1)d]=p2q2{{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}

⇒2a1+(p−1)d2a1+(p−1)d=pq\Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}

a1+(p−12)da1+(q−12)d=pq{{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}

For a6a21,  p=11, q=41{{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41

⇒a6a21=1141\Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}

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