項(xiàng)數(shù)為奇數(shù)的等差數(shù)列,奇數(shù)項(xiàng)之和為44,偶數(shù)項(xiàng)之和為33,求項(xiàng)數(shù)n及其中間一項(xiàng).

 

答案:
解析:

解:設(shè)數(shù)列共有2n+1項(xiàng),首項(xiàng)為a1,公差為d,其奇數(shù)項(xiàng)有n+1項(xiàng),偶數(shù)項(xiàng)有n項(xiàng),中間一項(xiàng)是第n+1項(xiàng),則有

 

n=3,

∴數(shù)列共有2n+1=2×3+1=7項(xiàng),其中第四項(xiàng)為中間一項(xiàng).

 


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