Question

Let $ {{a}_{1}},{{a}_{2}},{{a}_{3}},.... $ be terms of an A.P. If $ \dfrac{\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}...{{a}_{p}} \right)}{\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}...{{a}_{q}} \right)}=\dfrac{{{p}^{2}}}{{{q}^{2}}} $ , p not equal to q, then $ \dfrac{{{a}_{6}}}{{{a}_{21}}} $ equals
a. $ \dfrac{7}{2} $
b. $ \dfrac{2}{7} $
c. $ \dfrac{11}{41} $
d. $ \dfrac{41}{11} $

Answer 1

Hint: To deal with the type of question we will use the formula of sum of n terms of A.P. and finding the $ {{n}^{th}} $ term, which is $ \dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) $ and $ \left( a+\left( n-1 \right)d \right) $ respectively. In which ‘n’ is the number of terms, ‘a’ is the first term and ‘d’ is the difference of an A.P.

Complete step by step answer:
Moving ahead with the question in step wise manner, we had to find the value of $ \dfrac{{{a}_{6}}}{{{a}_{21}}} $ so as we know that $ {{a}_{n}} $ means $ {{n}^{th}} $ term of an A.P. means we can say that $ {{a}_{6}} $ and $ {{a}_{21}} $ are $ {{6}^{th}} $ and $ {{21}^{th}} $ term of an A.P. and as we also know that $ {{n}^{th}} $ term of an A.P. is $ \left( a+\left( n-1 \right)d \right) $ , so we can say that $ {{6}^{th}} $ and $ {{21}^{th}} $ term will be
 $ \begin{align}
  & {{a}_{1}}+\left( 6-1 \right)d \\
 & {{a}_{1}}+5d \\
\end{align} $ and $ \begin{align}
  & {{a}_{1}}+\left( 21-1 \right)d \\
 & {{a}_{1}}+20d \\
\end{align} $
So from here we can say that we need to find out the value of $ \dfrac{{{a}_{6}}}{{{a}_{21}}} $ which can be written as\[\dfrac{{{a}_{1}}+5d}{{{a}_{1}}+20d}\], means we if we get the value of ‘a’ and ‘d’ or relation among them then we will get the value. So let us use the given condition to find the relation or value of ‘a’ and ‘d’.
So we have $ \dfrac{\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}...{{a}_{p}} \right)}{\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}...{{a}_{q}} \right)}=\dfrac{{{p}^{2}}}{{{q}^{2}}} $ , in LHS side in numerator and denominator it represents the sum of terms starting from 1st term to $ {{p}^{th}} $ and $ {{q}^{th}} $ term, which we can write it as, sum of terms of A.P. starting from 1st term to $ {{p}^{th}} $ is $ \dfrac{p}{2}\left( 2{{a}_{1}}+\left( p-1 \right)d \right) $ similarly we can write for denominator series, which will be\[\dfrac{q}{2}\left( 2{{a}_{1}}+\left( q-1 \right)d \right)\]. So we can replace the series with this formula, which will give us;
 $ \dfrac{\dfrac{p}{2}\left( 2{{a}_{1}}+\left( p-1 \right)d \right)}{\dfrac{q}{2}\left( 2{{a}_{1}}+\left( q-1 \right)d \right)}=\dfrac{{{p}^{2}}}{{{q}^{2}}} $
On further simplifying it we will get;
 $ \begin{align}
  & \dfrac{p\left( 2{{a}_{1}}+\left( p-1 \right)d \right)}{q\left( 2{{a}_{1}}+\left( q-1 \right)d \right)}=\dfrac{{{p}^{2}}}{{{q}^{2}}} \\
 & \dfrac{2{{a}_{1}}+\left( p-1 \right)d}{2{{a}_{1}}+\left( q-1 \right)d}=\dfrac{p}{q} \\
\end{align} $
By cross multiplying we will get;
 $ \begin{align}
  & 2{{a}_{1}}q+qpd-qd=2{{a}_{1}}p+qpd-pd \\
 & 2{{a}_{1}}q+qpd-qd-2{{a}_{1}}p-qpd+pd=0 \\
 & 2{{a}_{1}}\left( q-p \right)-\left( q-p \right)d=0 \\
\end{align} $
By taking the $ q-p $ common we will get;
 $ \left( q-p \right)\left( 2{{a}_{1}}-d \right)=0 $
So from here we can say that either $ \left( q-p \right) $ is equal to zero, or $ \left( 2{{a}_{1}}-d \right) $ equal to zero. So by putting $ \left( 2{{a}_{1}}-d \right) $ equal to zero we will get;
 $ \begin{align}
  & 2{{a}_{1}}-d=0 \\
 & 2{{a}_{1}}=d \\
\end{align} $
So we got relation i.e. $ 2{{a}_{1}}=d $
Now put the value of ‘d’ in the expression whose we need to find the value which is;
\[\begin{align}
  & =\dfrac{{{a}_{6}}}{{{a}_{21}}} \\
 & =\dfrac{{{a}_{1}}+5d}{{{a}_{1}}+20d} \\
 & =\dfrac{{{a}_{1}}+5\left( 2{{a}_{1}} \right)}{{{a}_{1}}+20\left( 2{{a}_{1}} \right)} \\
\end{align}\]
On simplifying we get;
 $ \begin{align}
  & =\dfrac{11{{a}_{1}}}{41{{a}_{1}}} \\
 & =\dfrac{11}{41} \\
\end{align} $

So, the correct answer is “Option c”.

Note: The two conditions we got by solving the given condition which are $ \left( q-p \right) $ and $ \left( 2{{a}_{1}}-d \right) $ equal to zero. Then $ \left( q-p \right) $ will not be possible because on solving we will get $ q-p=0 $ which will give us $ q=p $ and according to the question ‘q’ is not equal to ‘p’.