A) 0
B) \[1\]
C) \[\frac{{{a}_{1}}}{{{a}_{2}}}\]
D) \[\frac{{{a}_{25}}}{{{a}_{24}}}\]
E) \[\frac{2{{a}_{1}}}{3{{a}_{2}}}\]
Correct Answer: C
Solution :
\[{{a}_{1}},{{a}_{2}},......,{{a}_{50}}\]are in GP. Let common ratio\[=r\] \[\Rightarrow \] \[{{a}_{2}}=r{{a}_{n}},{{a}_{3}}={{r}^{2}}{{a}_{1}},......,\]so on \[\therefore \] \[\frac{{{a}_{1}}-{{a}_{3}}+{{a}_{5}}-....+{{a}_{49}}}{{{a}_{2}}+{{a}_{4}}+{{a}_{6}}-.....+{{a}_{50}}}\] \[=\frac{{{a}_{1}}-{{r}^{2}}{{a}_{1}}+{{r}^{4}}{{a}_{1}}-....+{{r}^{48}}{{a}_{1}}}{{{a}_{1}}r-{{r}^{3}}{{a}_{1}}+{{r}^{5}}{{a}_{1}}-....+{{r}^{49}}{{a}_{1}}}\] \[=\frac{{{a}_{1}}(1-{{r}^{2}}+{{r}^{4}}-....+{{r}^{48}})}{{{a}_{1}}r(1-{{r}^{2}}+{{r}^{4}}-....+{{r}^{48}})}\] \[=\frac{1}{r}=\frac{{{a}_{1}}}{{{a}_{2}}}\]
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