Question

等差数列{an}中a1=9,a3+a8=0 求通项公式,

Answer 1

设公差为d

a1=a1

a2=a1+d

a3=a2+d=a1+2d

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an=a1+(n-1)d

a3+a8=a1+2d+a1+7d=0 --> d=-2

an= a1+(n-1)d

= 9+(n-1)(-2)

= 9-2(n-1)

= 11-2n