Question

已知4個(gè)數(shù)成等差數(shù)列,它們的和為26,中間兩項(xiàng)的積為40,求這4個(gè)數(shù).

Answer 1

解法一:設(shè)這4個(gè)數(shù)分別為a、b、c、d,

則

解之得或

所以這4個(gè)數(shù)為2,5,8,11或11,8,5,2.

解法二:設(shè)此等差數(shù)列的首項(xiàng)為a₁,公差為d,則

解之得或

所以這4個(gè)數(shù)為2,5,8,11或11,8,5,2.

解法三:設(shè)這4個(gè)數(shù)為a－3d,a－d,a+d,a+3d,則

??

所以這4個(gè)數(shù)為2,5,8,11或11,8,5,2.

點(diǎn)評(píng):等差數(shù)列中項(xiàng)的設(shè)法一般有兩種:

(1)通項(xiàng)法:設(shè)數(shù)列的通項(xiàng)公式,即設(shè)a_n=a₁+(n－1)d.

(2)對稱法:當(dāng)?shù)炔顢?shù)列{a_n}的項(xiàng)數(shù)n為奇數(shù)時(shí),可設(shè)中間的一項(xiàng)為a,再以公差為d向兩邊分別設(shè)項(xiàng):…,a－2d,a－d,a,a+d,a+2d,…;當(dāng)項(xiàng)數(shù)n為偶數(shù)時(shí),可設(shè)中間兩項(xiàng)分別為a－d,a+d,再以公差為2d向兩邊分別設(shè)項(xiàng):…,a－3d,a－d,a+d,a+3d,….