Question

在等差数列{an}中，公差d=2且a1 a2 a4成等比数列

(1)求a1的值 (2)求等差数列{an}的前n项和Sn 在线等

Answer 1

（1）因为a1，a2，a4成等比数列，所以（a2）^2=a1*a4,==> (a1+2)^2=a1(a1+6)==> a1^2+4a1+4=a1^2+6a1 ==> a1^2+4a1+4-(a1^2+6a1)=0==> -2a1+4=0,==>2a1=4,==>a1=2

(2)an=2+2(n-1)=2n,
sn=2n+n^2-n=n^2+n

Answer 2

a2*a2=a1*a4
a2=a1+d
a4=a1+3d
所以可以得出a1=2
Sn=na1+n(n-1)d/2=n*n+n

Answer 3

(1)∵an是等差数列,d=2且a1 a2 a4成等比数列,
∴a2^2=a1*a4 即 (a1+2)^2=a1*(a1+6) 解得a1=2
即an=a1+(n-1)d=2+2(n-1)=2n

(2)sn=(a1+an)n/2=(2+2n)n/2=n+n^2

Answer 4

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