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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
Let ana_n be the nnth term of an AP. If ∑r=11099a2r=10100\sum\limits_{{r=1}}^{10^{99}} a_{2 r}=10^{100} and ∑r=11099a2r−1=1099\sum\limits_{{r=1}}^{10^{99}} a_{2 r-1}=10^{99}, then the common difference of an AP is
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Correct option: B
9
Option B (9) is correct because the common difference of the AP is calculated to be 9 based on the given summations.

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