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Let a_{1},a_{2},a_{3}........ be an A.P. such that \frac{a_{1}+a_{2}+a_{3}......+a_{p}}{a_{1}+a_{2}+a_{3}......+a_{q}}=\frac{p^{3}}{q^{3}};p\neq q .

Then \frac{a_{6}}{a_{21}} is equal to :

Option: 1

\frac{41}{11}


Option: 2

\frac{121}{1681}


Option: 3

\frac{11}{41}


Option: 4

\frac{121}{1681}


Answers (1)

best_answer

As we learnt in 

Sum of n terms of an AP -

S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]

and

Sum of n terms of an AP

S_{n}= \frac{n}{2}\left [ a+l\right ]

 

Given a1 ,a2 ,a3...are in A.P and

 \frac{a_{1}+a_{2}+a_{3}+......+a_{p}}{a_{1}+a_{2}+a_{3}+....+a_{q}}=\frac{p^{3}}{q^{3}}

\therefore \frac{\frac{p}{2}[2a_{1}+(p-1)d]}{\frac{q}{2}[2a_{1}+(q-1)d]}=\frac{p^{3}}{q^{3}}

\Rightarrow \frac{[2a_{1}+(p-1)d]}{[2a_{1}+(q-1)d]}=\frac {p^2}{q^2}

\Rightarrow \frac{[a_{1}+\frac{(p-1)}{2}d]}{[a_{1}+\frac{(q-1)}{2}d]}=\frac {p^2}{q^2}

To get ratio of 6th and 21st terms, we can put p = 11 and q = 41 in this equation

\Rightarrow \frac{[a_{1}+5d]}{[a_{1}+21d]}=\frac {11^2}{41^2}

\Rightarrow \frac{[a_{1}+5d]}{[a_{1}+21d]}=\frac{a_6}{a_{21}}=\frac {11^2}{41^2}=\frac{121}{1681}

 

Posted by

Suraj Bhandari

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