等差数列AN中D=2 ,AN=11,SN=35 求A1与N?

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等差数列AN中D=2,AN=11,SN=35求A1与N?等差数列AN中D=2,AN=11,SN=35求A1与N?等差数列AN中D=2,AN=11,SN=35求A1与N?An=A1+(n-1)d=A1+

等差数列AN中D=2 ,AN=11,SN=35 求A1与N?
等差数列AN中D=2 ,AN=11,SN=35 求A1与N?

等差数列AN中D=2 ,AN=11,SN=35 求A1与N?
An=A1+(n-1)d=A1+2(n-1)=11,
Sn=nA1+n(n-1)d/2=nA1+n(n-1)=35,
解方程组:
A1+2(n-1)=11
nA1+n(n-1)=35;
A1= 3,n=5,
A1=-1,n=7,

Sn=nA1+n(n-1)d/2,再看看这个式子:Sn=nAn-n(n-1)d/2,把D=2 ,AN=11,SN=35 代入就解出来了。。

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