有这样一类数,a1\a2\a3\.满足公式an=a1+[n-1]d已知a2=97,a5=85,问【1】a1和d的值!【2】若ak>0,ak+1<0,求k的值
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![有这样一类数,a1\a2\a3\.满足公式an=a1+[n-1]d已知a2=97,a5=85,问【1】a1和d的值!【2】若ak>0,ak+1<0,求k的值](/uploads/image/z/10440524-20-4.jpg?t=%E6%9C%89%E8%BF%99%E6%A0%B7%E4%B8%80%E7%B1%BB%E6%95%B0%2Ca1%5Ca2%5Ca3%5C.%E6%BB%A1%E8%B6%B3%E5%85%AC%E5%BC%8Fan%3Da1%2B%5Bn-1%5Dd%E5%B7%B2%E7%9F%A5a2%3D97%2Ca5%3D85%2C%E9%97%AE%E3%80%901%E3%80%91a1%E5%92%8Cd%E7%9A%84%E5%80%BC%21%E3%80%902%E3%80%91%E8%8B%A5ak%EF%BC%9E0%2Cak%2B1%EF%BC%9C0%2C%E6%B1%82k%E7%9A%84%E5%80%BC)
有这样一类数,a1\a2\a3\.满足公式an=a1+[n-1]d已知a2=97,a5=85,问【1】a1和d的值!【2】若ak>0,ak+1<0,求k的值
有这样一类数,a1\a2\a3\.满足公式an=a1+[n-1]d已知a2=97,a5=85,问【1】a1和d的值!
【2】若ak>0,ak+1<0,求k的值
有这样一类数,a1\a2\a3\.满足公式an=a1+[n-1]d已知a2=97,a5=85,问【1】a1和d的值!【2】若ak>0,ak+1<0,求k的值
[1]由题意可知:a2=a1+d=97;
a5=a1+4d=85;
由以上两式可得:a1=101; d= -4.
[2]由第一问可知;an=a1+(n-1)d=105-4n;
因为ak>0,ak+10且105-4(k+1)
a1+d=97
a1+4d=85
解得a1=101,d=-4
101-4(k-1)>0,101-4(k+1-1)<0
解得25.25
a2=a1+(2-1)d=97
a5=a1+(5-1)d=85
d=-4 a1=93
另外 第二问是不是 错了 ???