Let a1,a2,a3,.... be terms of an A.P. If a1+a2+…..apa1+a2+……aq=p2q2,p≠q then a6a21 equals

p22a1+(p−1)dq22a1+(q−1)d=p2q2⇒q2a1+(p−1)d =p(2a1+(q-1)d) ⇒2a1=da6a21=a1+5da1+20d=a1+10a1a1+40a1=1141

Let a1,a2,a3,.... be terms of an A.P. If a1+a2+…..apa1+a2+……aq=p2q2,p≠q then a6a21 equals

p22a1+(p−1)dq22a1+(q−1)d=p2q2⇒q2a1+(p−1)d =p(2a1+(q-1)d) ⇒2a1=da6a21=a1+5da1+20d=a1+10a1a1+40a1=1141

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Let a_1,a_2,a_3,.... be terms of an A.P. If $\frac{a_{1} + a_{2} + \dots . . a_{p}}{a_{1} + a_{2} + \dots \dots a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q$ then $\frac{a_{6}}{a_{21}}$ equals

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41/11

7/2

2/7

11/41

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{\frac{p}{2} (2 a_{1} + (p - 1) d)}{\frac{q}{2} (2 a_{1} + (q - 1) d)} = \frac{p^{2}}{q^{2}} \Rightarrow q (2 a_{1} + (p - 1) d) = p (2 a_{1} + (q - 1) d) \\ \Rightarrow 2 a_{1} = d \\ \frac{a_{6}}{a_{21}} = \frac{a_{1} + 5 d}{a_{1} + 20 d} = \frac{a_{1} + 10 a_{1}}{a_{1} + 40 a_{1}} = \frac{11}{41} \end{array}$