A) \[1\]
B) \[9\]
C) \[10\]
D) \[{{10}^{99}}\]
Correct Answer: B
Solution :
Let\[d\]be the common difference of the A.P. then \[{{a}_{2r}}={{a}_{2r-1}}+d\] \[\Rightarrow \] \[\sum\limits_{r\,=\,1}^{{{10}^{99}}}{{{a}_{2r}}}=\sum\limits_{r\,=\,1}^{{{10}^{99}}}{({{a}_{2r-1}}+d)}\] \[\Rightarrow \] \[\sum\limits_{r\,=\,1}^{{{10}^{99}}}{{{a}_{2r}}}=\sum\limits_{r\,=\,1}^{{{10}^{99}}}{{{a}_{2r-1}}}+{{10}^{99}}d\] \[\Rightarrow \] \[{{10}^{100}}={{10}^{99}}+{{10}^{99}}d\] \[\Rightarrow \] \[d=\frac{{{10}^{100}}-{{10}^{99}}}{{{10}^{99}}}=\frac{{{10}^{99}}(10-1)}{{{10}^{99}}}=9\]
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