A) 300
B) 150
C) 175
D) 225
Correct Answer: B
Solution :
\[\sum\limits_{n=1}^{100}{{{a}_{2n+1}}=\frac{{{a}_{1}}{{r}^{2}}\left( {{r}^{200}}-1 \right)}{{{r}^{2}}-1}}=200\] ...(i) and \[\sum\limits_{n=1}^{100}{{{a}_{2n}}=\frac{{{a}_{1}}r\left( {{r}^{200}}-1 \right)}{{{r}^{2}}-1}}=100\] ...(i) Dividing (i) by (ii) we get \[r=2\] and adding (i) and (ii) we get \[{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}+....+{{a}_{200}}+{{a}_{201}}=300\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{a}_{1}}r+{{a}_{2}}r+{{a}_{3}}r+....+{{a}_{200}}r=300\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r({{a}_{1}}+{{a}_{2}}+.....+{{a}_{200}})=300\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{a}_{1}}+{{a}_{2}}+.....+{{a}_{200}}=150\]You need to login to perform this action.
You will be redirected in
3 sec