问题
填空题
在等差数列{an}中,公差d≠0,a2,a4,a7,成等比数列,则
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答案
∵a2,a4,a7成等比数列,
∴a2•a7=a42,即(a1+d)(a1+6d)=(a1+3d)2,
解得a1=3d,或d=0(舍去),
由等差数列通项公式得an=a1+(n-1)d=3d+(n-1)d=(n+2)d
故
=a1+a4 a2
=3d+6d 4d
.9 4
故答案为:9 4
在等差数列{an}中,公差d≠0,a2,a4,a7,成等比数列,则
|
∵a2,a4,a7成等比数列,
∴a2•a7=a42,即(a1+d)(a1+6d)=(a1+3d)2,
解得a1=3d,或d=0(舍去),
由等差数列通项公式得an=a1+(n-1)d=3d+(n-1)d=(n+2)d
故
=a1+a4 a2
=3d+6d 4d
.9 4
故答案为:9 4