设公比为q.由已知条件知a1,a3,a11成等比数列.
a3²=a1×a11
(a1+2d)²=a1(a1+10d)
整理,得
2d²-3a1d=0
d(2d-3a1)=0
d=0(与已知矛盾,舍去)或d=3a1/2
a3=a1+2d=a1+3a1=4a1
q=a3/a1=4a1/a1=4
a(k6)=a(k1)q^5
=a1q^5
=a1×4^5
=1024a1
=a1+1023a1
=a1+1023×(2/3)d
=a1+682d
=a1+(683-1)d
k6=683
设公比为q.由已知条件知a1,a3,a11成等比数列.
a3²=a1×a11
(a1+2d)²=a1(a1+10d)
整理,得
2d²-3a1d=0
d(2d-3a1)=0
d=0(与已知矛盾,舍去)或d=3a1/2
a3=a1+2d=a1+3a1=4a1
q=a3/a1=4a1/a1=4
a(k6)=a(k1)q^5
=a1q^5
=a1×4^5
=1024a1
=a1+1023a1
=a1+1023×(2/3)d
=a1+682d
=a1+(683-1)d
k6=683