已知5个数成等差数列.它们的和为5.平方和为.求这5个数解:设三个数为a.公差为d.则这5个数依次为a-2d.a-d ,a ,a+d ,a+2d依题意: 2 +(a-d)2 + a2 + (a——青夏教育精英家教网—

已知5个数成等差数列.它们的和为5.平方和为.求这5个数解:设三个数为a.公差为d.则这5个数依次为a-2d.a-d ,a ,a+d ,a+2d依题意: 2 +(a-d)2 + a2 + (a+d)2 + 2 = 且 + a + = 5 即 a2+2d2 = 且 a=1 ∴a=1且d= 当d=时.这5个数分别是-..1.., 当d=-时.这5个数分别是..1..- 设数列..满足:.(n=1,2,3,-).证明为等差数列的充分必要条件是为等差数列且(n=1,2,3,-) 证明:必要性．设{a n}是公差为 d1的等差数列.则 b n+1b n = (a n+1a n+3)(a na n+2)=(a n+1a n)(a n+3a n+2)=d1d1=0 所以b n≤b n+1 成立. 又c n+1-c n=(a n+1-a n)+2(a n+2-a n+1)+3(a n+3-a n+2) =d1+2d1+3d1=6d1. 所以数列{c n}为等差数列. 充分性.设数列{c n}是公差为d2的等差数列.且b n≤b n+1 . 证法一: ∵c n= a n +2a n+1+3a n+2 ① ∴c n+2= a n+2+2a n+3+3a n+4 ② ①-②得c n - c n+2=( a n - a n+2)+2(a n+1 - a n+3)+3(a n+2 - a n+4) = b n + 2b n+1 + 3b n+2. ∵c n- c n+2=( c n- c n+1)+( c n+1 - c n+2)=-2d2. ∴bn + 2bn+1 + 3bn+2 =-2d2. ③ 从而有 bn+1 + 2bn+2 + 3bn+3 =-2d2. ④ ④-③得 (b n+1 - b n)+2(b n+2 - b n+1)+3(b n+3 - b n+2)=0. ⑤ ∵b n+1 - b n≥0.b n+2 - b n+1≥0, b n+3 - b n+2≥0. ∴由⑤得b n+1 - b n=0. 由此不妨设b n =d3.则a n - a n+2 =d3. 由此c n = a n + 2a n+1 + 3a n+2 = 4a n + 2a n+1 – 3d3, 从而c n+1 = 4a n+1 + 2a n+2 - 3d3 = 4a n+1 + 2a n -5d3. 两式相减得c n+1 - c n =2(a n+1 - a n)-2d3, 因此a n+1 - a n =(c n+1-cn)+d3=d2+d3, 所以数列{a n}是等差数列. 证法二: 令An = a n+1- a n.由b n≤b n+1知a n - a n+2≤a n+1- a n+3, 从而a n+1- a n≥a n+3 - a n+2,即An≥An+2. 由c n = a n + 2a n+1 + 3a n+2, c n+1 = a n+1 + 2a n+2 + 3a n+3得 c n+1-c n=( a n+1- a n)+2(a n+2- a n+1)+3(a n+3 - a n+2).即 An+2An+1+3An+2=d2. ⑥ 由此得 An+2+2An+3+3An+4=d2. ⑦ ⑥-⑦得 (An-An+2)+2(An+1- An+3)+3(An+2- An+4)=0. ⑧ 因为An-An+2≥0.An+1- An+3≥0.An+2- An+4≥0, 所以由⑧得An-An+2=0. 于是由⑥得 4An+2An+1=An+2An+1+3An+2=d2, ⑨ 从而 2An+4An+1=4An+1+2An+2=d2. ⑩ 由⑨和⑩得4An+2An+1=2An+4An+1,故An+1= An ,即 a n+2- a n+1= a n+1- a n. 所以数列{a n}是等差数列. 【查看更多】

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已知5个数成等差数列，它们的和为5，平方和为，求这5个数