已知数列{an}是由正数组成的等差数列,Sn是其前n项和,并且a3=5,a4S2=28

(1)设公差为d,则a4=a3+d=5+d,S2=a1+a2=(a3-2d)+(a3-d)=2a3-3d=10-3d
所以(5+d)(10-3d)=28,解得d=2或d=-11/3
因为数列{an}是由正数组成,所以d>0,所以d=2
(2)a1=a3-2d=5-2*2=1,所以an=2n-1
(1+1/a1)(1+2/a2)(1+3/a3)...(1+1/an)
=2/1*4/3*6/5*...*2n/(2n-1)
因为2/1>3/2,4/3>5/4,2n/(2n-1)>(2n+1)/2n
所以[2/1*4/3*6/5*...*2n/(2n-1)]^2
>2/1*3/2*4/3*5/4...2n/(2n-1)*(2n+1)/2n
=2n+1
所以2/1*4/3*6/5*...*2n/(2n-1)>根号(2n+1)>=a根号(2n+1)
所以a

a3=a1+2d=5 ==> a1=5-2d
a4s2=(a1+3d)(a1+a2)=(a1+3d)(2a1+d)=28
=> (5+d)(10-3d)=28
=> 3d^2+5d-22=0
=> (3d+11)(d-2)=0
d1=-11/3 d2=2
又"已知数列{an}是由正数组成的等差数列"
所以 d=2 => a1=...

因为已知是等差数列,设公差为d，a3=a1+2d=5 a1=5-2d
根据a4S2=28
可推出(a1+3d)(2a1+d)=28
将a1带入:(5+d)(10-3d)=28
打开括号，整理：3d~2+5d-22=0