设{an}是一个以d(d>o)的等差数列,若1/a1a2+1/a2a3=1/a3a4=3/4,且前6项和s6=21,则a
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1/a1a2=(1/d)(1/a1 - 1/a2),

1/a2a3=(1/d)(1/a2 - 1/a3),

1/a3a4=(1/d)(1/a3 - 1/a4),

则由 1/a1a2+1/a2a3=1/a3a4 得

(1/d)[(1/a1 - 1/a2)+(1/a2 - 1/a3)]=(1/d)(1/a1 - 1/a3)

=(1/d)(1/a3 - 1/a4)

则1/a1 + 1/a4 =2/a3

→a3(a1+a4)=2a1·a4

a3·[(a3-2d)+(a3+d)]=2(a3-2d)·(a3+d)

a3·(2a3-d)=2(a3^2 -a3·d -2d^2)

整理得

a3 = -4d

由前6项和s6=21得

s6=3(a3+a4)=3(2a3+d)=3×(-7d)=-21d

即d=1

a3=-4;

a1=a3-2d=-6;

an=-6n+(n-1)=-5n-1

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