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Sequences and Series question

2006 · Shift 0 · Q75
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Sequences and Series question

2006 · Shift 0 · Q75

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let a1{a_1}a1​, a2{a_2}a2​, a3{a_3}a3​.....be terms on A.P. If a1+a2+.....apa1+a2+.....aq=p2q2, peq, then a6a21 {{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p e q,\,then\,{{{a_6}} \over {{a_{21}}}}\,a1​+a2​+.....aq​a1​+a2​+.....ap​​=q2p2​,peq,thena21​a6​​ equals
  1. A
    4111{{41} \over {11}}1141​
  2. B
    72{7 \over 2}27​
  3. C
    27{2 \over 7}72​
  4. D
    1141{{11} \over {41}}4111​
View written solutionFree

Correct answer: D

  1. Let the A.P. have first term aaa and common difference ddd.

    Then the sum of first nnn terms is Sn=n2[2a+(n−1)d].S_n=\frac{n}{2}\left[2a+(n-1)d\right].Sn​=2n​[2a+(n−1)d].

  2. Given SpSq=p2q2,p≠q.\frac{S_p}{S_q}=\frac{p^2}{q^2}, \qquad p\ne q.Sq​Sp​​=q2p2​,p=q.

    Substitute the formula for sums: p2[2a+(p−1)d]q2[2a+(q−1)d]=p2q2.\frac{\frac{p}{2}[2a+(p-1)d]}{\frac{q}{2}[2a+(q-1)d]}=\frac{p^2}{q^2}.2q​[2a+(q−1)d]2p​[2a+(p−1)d]​=q2p2​.

    Simplifying, p[2a+(p−1)d]q[2a+(q−1)d]=p2q2.\frac{p[2a+(p-1)d]}{q[2a+(q-1)d]}=\frac{p^2}{q^2}.q[2a+(q−1)d]p[2a+(p−1)d]​=q2p2​.

    Cross-multiply: q[2a+(p−1)d]=p[2a+(q−1)d].q[2a+(p-1)d]=p[2a+(q-1)d].q[2a+(p−1)d]=p[2a+(q−1)d].

  3. Expand both sides: 2aq+q(p−1)d=2ap+p(q−1)d.2aq+q(p-1)d=2ap+p(q-1)d.2aq+q(p−1)d=2ap+p(q−1)d.

    Rearranging, 2a(q−p)+(qp−q−pq+p)d=0.2a(q-p)+\big(qp-q-pq+p\big)d=0.2a(q−p)+(qp−q−pq+p)d=0.

    2a(q−p)+(p−q)d=0.2a(q-p)+(p-q)d=0.2a(q−p)+(p−q)d=0.

    (q−p)(2a−d)=0. (q-p)(2a-d)=0.(q−p)(2a−d)=0.

    Since p≠qp\ne qp=q, we must have 2a−d=0⇒d=2a.2a-d=0 \quad\Rightarrow\quad d=2a.2a−d=0⇒d=2a.

  4. Therefore the general term is an=a+(n−1)d=a+(n−1)(2a)=a(2n−1).a_n=a+(n-1)d=a+(n-1)(2a)=a(2n-1).an​=a+(n−1)d=a+(n−1)(2a)=a(2n−1).

  5. Now compute a6a_6a6​ and a21a_{21}a21​: a6=a(2⋅6−1)=11a,a_6=a(2\cdot 6-1)=11a,a6​=a(2⋅6−1)=11a, a21=a(2⋅21−1)=41a.a_{21}=a(2\cdot 21-1)=41a.a21​=a(2⋅21−1)=41a.

    Hence, a6a21=11a41a=1141.\frac{a_6}{a_{21}}=\frac{11a}{41a}=\frac{11}{41}.a21​a6​​=41a11a​=4111​.

  6. So the correct option is 1141.\boxed{\frac{11}{41}}.4111​​.

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