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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
If {ai}i=1n,\left\{ {{a_i}} \right\}_{i = 1}^n, where nn is an even integer, is an arithmetic progression with common difference 1, and ∑i=1nai=192,  ∑i=1n/2a2i=120,\sum\limits_{i = 1}^n {{a_i}} = 192,\,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}}} = 120, then nn is equal to:
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Correct option: B
96
Option B (96) is correct because by setting up two equations based on the given information about the arithmetic progression and solving them simultaneously, we find that n = 96.

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