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Let  a_1,a_2,a_3...  be terms of an A.P.   If   \frac{a_1+a_2+...+a_p}{a_1+a_2+...+a_q}=\frac{p^2}{q^2}     , p\ne q   then the value of  \frac{a_6}{a_{21}}  is 

A.  \frac{41}{11}                        B.   \frac{7}{2}                       C.  \frac{2}{7}                         D.  \frac{11}{41}

Answers (2)

@jeetu

s_n=\frac{n}{2}(a_1+a_n)\\given\frac{\frac{p(a_1+a_p)}{2}}{\frac{q(a_1+a_q)}{2}}=\frac{p^2}{q^2}\\\Rightarrow \frac{a_1+a_p}{a_1+a_q}=\frac{p}{q}\\\therefore \frac{a1+a_6}{a_1+a_21}=\frac{2}{7}\Rightarrow 5a_1=2a_{21}-7a_{6}\\\Rightarrow5a_1= 2(a_1+20d)-7(a_1+5d)\\\Rightarrow 10a_1=56\\\therefore \frac{a_6}{a_21}=\frac{a_1+5d}{a_1+20d}\\=\frac{11}{41}

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Safeer PP

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s_n=\frac{n}{2}(a_1+a_n)\\\Rightarrow \frac{a_1+a_6}{a_1+a_21}=\frac{2}{7}\\\Rightarrow \frac{a_1+a_1+5d}{a_1+a_1+20d}=\frac{2}{7}\\\Rightarrow 10a_1=5d\\\therefore \frac{a_6}{a_{21}}=\frac{11}{41}

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Safeer PP

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