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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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an=?
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
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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