MathematicsMedium178×since 2002Q5225If {ai}i=1n\{ {a_i}\} _{i = 1}^n{ai}i=1n, where n 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}i=1∑nai=192,i=1∑n/2a2i=120, then n is equal to :A48B96C92D104Check answerSkip