A) \[\frac{41}{11}\]
B) \[\frac{121}{1681}\]
C) \[\frac{11}{41}\]
D) \[\frac{121}{1861}\]
Correct Answer: D
Solution :
Given that \[\frac{{{a}_{1}}+{{a}_{2}}+....{{a}_{p}}}{{{a}_{1}}+{{a}_{2}}+....+{{a}_{r}}}=\frac{{{p}^{3}}}{{{q}^{3}}}\] \[\Rightarrow \] \[\frac{\frac{p}{2}[2{{a}_{1}}+(p-1)d]}{\frac{q}{2}[2{{a}_{1}}+(q-1)d]}=\frac{{{p}^{3}}}{{{q}^{3}}}\] When d be a common difference of an AP \[\frac{{{a}_{1}}+\left( \frac{p-1}{2} \right)d}{{{a}_{1}}+\left( \frac{q-1}{2} \right)d}=\frac{{{p}^{2}}}{{{q}^{2}}}\] \[\because \] \[\frac{{{a}_{6}}}{{{a}_{21}}}=\frac{{{a}_{1}}+5d}{{{a}_{1}}+20d}\] \[\therefore \] \[\frac{p-1}{2}=5\] \[\Rightarrow \] \[p=11\] \[\frac{q-1}{2}=20\] \[\Rightarrow \] \[q=41\] \[\therefore \] \[\frac{{{a}_{1}}+5d}{{{a}_{1}}+20d}=\frac{{{11}^{2}}}{{{41}^{2}}}=\frac{121}{1681}\]
You need to login to perform this action.
You will be redirected in
3 sec
