等差数列{an}中,a1+a6=0,a3a4=-1,则am= 求通项公式

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等差数列{an}中,a1+a6=0,a3a4=-1,则am= 求通项公式
等差数列{an}中,a1+a6=0,a3a4=-1,则am= 求通项公式

等差数列{an}中,a1+a6=0,a3a4=-1,则am= 求通项公式
a1+a6=a3+a4=0;
a3a4=-1;
∴a3=1,a4=-1;或a3=-1,a4=1;
∴d=-1-1=-2或d=1-(-1)=2;
∴an=a1+(n-1)d=5+(n-1)(-2)=8-2n或-5+2(n-1)=2n-7;
如果本题有什么不明白可以追问,

am=?????
{An}等差数列, 则a(n) = a1 + (n-1)d
a(3)*a(4) = (a1 + 2d)(a1 + 3d) = -1
a(1) + a(6) = a(1 ) + (a1 + 5d) = 0
求得;a1 = 5 d = - 2..............
或者 a1 = -5 d = 2...

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am=?????
{An}等差数列, 则a(n) = a1 + (n-1)d
a(3)*a(4) = (a1 + 2d)(a1 + 3d) = -1
a(1) + a(6) = a(1 ) + (a1 + 5d) = 0
求得;a1 = 5 d = - 2..............
或者 a1 = -5 d = 2..............
故 通项公式 A(n) = 5 -2 (n - 1) = -2n +7
A(n) =- 5 +2 (n - 1) = 2n -7

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