设{An}是公差不为零的等差数列,Sn为其前n项和,满足A2^2+A3^2=A4^2+A5^2,S7=7.
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![设{An}是公差不为零的等差数列,Sn为其前n项和,满足A2^2+A3^2=A4^2+A5^2,S7=7.](/uploads/image/z/2427681-57-1.jpg?t=%E8%AE%BE%7BAn%7D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%8D%E4%B8%BA%E9%9B%B6%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2CSn%E4%B8%BA%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%BB%A1%E8%B6%B3A2%5E2%2BA3%5E2%3DA4%5E2%2BA5%5E2%2CS7%3D7.)
设{An}是公差不为零的等差数列,Sn为其前n项和,满足A2^2+A3^2=A4^2+A5^2,S7=7.
设{An}是公差不为零的等差数列,Sn为其前n项和,满足A2^2+A3^2=A4^2+A5^2,S7=7.
设{An}是公差不为零的等差数列,Sn为其前n项和,满足A2^2+A3^2=A4^2+A5^2,S7=7.
由条件
(a1+d)²+(a1+2d)²=(a1+3d)²+(a1+4d)²
a1²+2a1d+d²+a1²+4a1d+4d²=a1²+6a1d+9d²+a1²+8a1d+16d²
2a1²+6a1d+5d²=2a1²+14a1d+25d²
8a1d+20d²=0
2a1d+5d²=0
d(2a1+5d)=0
d=0,或者2a1+5d=0 (1)
S7=(a1+a7)7/2=(a1+a1+6d)7/2=7(a1+3d)=7
a1+3d=1 (2)
当d=0时,代入上式,解得a1=1,得an=1
或者由(1)(2)
解得a1=-5,d=2得an=-5+2*(n-1)=2n-7
很高兴为您解答,本题有不理解的请追问!
公差为d d不为零
S7=(a1+a7)*7/2=a4*7=7
所以a4=1;
a2^2+a3^2=a4^2+a5^2
a5^2-a3^2=a2^2-a4^2
2d(a5+a3)=-2d(a2+a4)
(a5+a3)+(a2+a4)=0
2*a4+2*a3=0
所以a3=-1;
d=2
an=2n-7; S=n(n-6)
s7=a1+a2+a3+a4+a5+a6+a7=7a4=7 得a4=1 带入
a2²+a3²=a4²+a5²
(1-2d)²+(1-d)²=1+(1+d)²化简
d²-2d=0得d=2或d=0(舍弃)