If \{ a_{i} \}_{i = 1}^{n}, where n is an even integer, is an arithmetic progression with common difference 1 , and \sum_{i = 1}^{n}\mspace{2mu} a_{i} = 192,\sum_{i = 1}^{n/2}\mspace{2mu} a_{2i} = 120, and then n is equal to: Sequence & Series (Mathematics) | DivineJEE