A) \[\frac{7}{2}\]
B) \[\frac{2}{7}\]
C) \[\frac{11}{41}\]
D) \[\frac{41}{11}\]
Correct Answer: C
Solution :
Use the formula,\[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\]and simplify it. Since, \[\frac{{{a}_{1}}+{{a}_{2}}+.....+{{a}_{p}}}{{{a}_{1}}+{{a}_{2}}+....{{a}_{p}}}=\frac{{{p}^{2}}}{{{q}^{2}}}\] \[\therefore \]\[\frac{\frac{p}{2}[2{{a}_{1}}+(p-1)d]}{\frac{q}{2}[2{{a}_{1}}+(q-1)d]}=\frac{{{p}^{2}}}{{{q}^{2}}}\] (\[\because \]d is a common, difference of an AP) \[\Rightarrow \]\[\frac{(2{{a}_{1}}-d)+pd}{(2{{a}_{1}}-d)+qd}=\frac{p}{q}\] \[\Rightarrow \]\[(2{{a}_{1}}-d)(p-q)=0\] \[\Rightarrow \]\[{{a}_{1}}=\frac{d}{2}\] Now. \[\frac{{{a}_{6}}}{{{a}_{21}}}=\frac{{{a}_{1}}+5d}{{{a}_{1}}+20d}=\frac{\frac{d}{2}+5d}{\frac{d}{2}+20d}=\frac{11d}{41d}=\frac{11}{41}\]
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