JEE Math Practice Question
If $$\{ {a_i}\} _{i = 1}^n$$, where n is an even integer, is an arithmetic progression with common difference 1, and $$\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120} $$, then n is equal to :
- A.48
- B.96
- C.92
- D.104
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