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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2012 · Mathematics · Sequences and Series

Let a_ 1 , a_ 2 , a_ 3 . be terms on A.P. If a_ 1 +a_ 2 + a_ p a_ 1...

Let a1,a2,a3………….a_{1}, a_{2}, a_{3} \ldots \ldots \ldots \ldots. be terms on A.P. If a1+a2+………apa1+a2+……….+aq=p2q2,p≠q,\frac{a_{1}+a_{2}+\ldots \ldots \ldots a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots.+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q, then a6a21\frac{a_{6}}{a_{21}} equals

Options

  1. A4111\frac{41}{11}
  2. B72\frac{7}{2}
  3. C27\frac{2}{7}
  4. D1141\frac{11}{41}

Correct answer

D. 1141\frac{11}{41}

Step-by-step solution

p2[2a1+(p−1)d]q2[2a1+(q−1)d]=p2q2⇒2a1+(p−1)d2a1+(q−1)d=pqa1+(p−12)da1+(q−12)d=pq\frac{\frac{p}{2}\left[2 a_{1}+(p-1) d\right]}{\frac{q}{2}\left[2 a_{1}+(q-1) d\right]}=\frac{p^{2}}{q^{2}}\Rightarrow \frac{2 a_{1}+(p-1) d}{2 a_{1}+(q-1) d}=\frac{p}{q}\frac{a_{1}+\left(\frac{p-1}{2}\right) d}{a_{1}+\left(\frac{q-1}{2}\right) d}=\frac{p}{q} For a6a21,p=11,q=41\frac{a_{6}}{a_{21}}, \quad p=11, q=41
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