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Question

Question: Let \( {{a}_{1}},{{a}_{2}},{{a}_{3}},.... \) be terms of an A.P. If \( \dfrac{\left( {{a}_{1}}+{{a}_...

Let a1,a2,a3,....{{a}_{1}},{{a}_{2}},{{a}_{3}},.... be terms of an A.P. If (a1+a2+a3...ap)(a1+a2+a3...aq)=p2q2\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 a6a21\dfrac{{{a}_{6}}}{{{a}_{21}}} equals
a. 72\dfrac{7}{2}
b. 27\dfrac{2}{7}
c. 1141\dfrac{11}{41}
d. 4111\dfrac{41}{11}

Explanation

Solution

To deal with the type of question we will use the formula of sum of n terms of A.P. and finding the nth{{n}^{th}} term, which is n2(2a+(n1)d)\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) and (a+(n1)d)\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 a6a21\dfrac{{{a}_{6}}}{{{a}_{21}}} so as we know that an{{a}_{n}} means nth{{n}^{th}} term of an A.P. means we can say that a6{{a}_{6}} and a21{{a}_{21}} are 6th{{6}^{th}} and 21th{{21}^{th}} term of an A.P. and as we also know that nth{{n}^{th}} term of an A.P. is (a+(n1)d)\left( a+\left( n-1 \right)d \right) , so we can say that 6th{{6}^{th}} and 21th{{21}^{th}} term will be
a1+(61)d a1+5d \begin{aligned} & {{a}_{1}}+\left( 6-1 \right)d \\\ & {{a}_{1}}+5d \\\ \end{aligned} and a1+(211)d a1+20d \begin{aligned} & {{a}_{1}}+\left( 21-1 \right)d \\\ & {{a}_{1}}+20d \\\ \end{aligned}
So from here we can say that we need to find out the value of a6a21\dfrac{{{a}_{6}}}{{{a}_{21}}} which can be written asa1+5da1+20d\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 (a1+a2+a3...ap)(a1+a2+a3...aq)=p2q2\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 pth{{p}^{th}} and qth{{q}^{th}} term, which we can write it as, sum of terms of A.P. starting from 1st term to pth{{p}^{th}} is p2(2a1+(p1)d)\dfrac{p}{2}\left( 2{{a}_{1}}+\left( p-1 \right)d \right) similarly we can write for denominator series, which will beq2(2a1+(q1)d)\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;
p2(2a1+(p1)d)q2(2a1+(q1)d)=p2q2\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;
p(2a1+(p1)d)q(2a1+(q1)d)=p2q2 2a1+(p1)d2a1+(q1)d=pq \begin{aligned} & \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{aligned}
By cross multiplying we will get;
2a1q+qpdqd=2a1p+qpdpd 2a1q+qpdqd2a1pqpd+pd=0 2a1(qp)(qp)d=0 \begin{aligned} & 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{aligned}
By taking the qpq-p common we will get;
(qp)(2a1d)=0\left( q-p \right)\left( 2{{a}_{1}}-d \right)=0
So from here we can say that either (qp)\left( q-p \right) is equal to zero, or (2a1d)\left( 2{{a}_{1}}-d \right) equal to zero. So by putting (2a1d)\left( 2{{a}_{1}}-d \right) equal to zero we will get;
2a1d=0 2a1=d \begin{aligned} & 2{{a}_{1}}-d=0 \\\ & 2{{a}_{1}}=d \\\ \end{aligned}
So we got relation i.e. 2a1=d2{{a}_{1}}=d
Now put the value of ‘d’ in the expression whose we need to find the value which is;

& =\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{aligned}$$ On simplifying we get; $ \begin{aligned} & =\dfrac{11{{a}_{1}}}{41{{a}_{1}}} \\\ & =\dfrac{11}{41} \\\ \end{aligned} $ **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’.