1.S2= a1+a2 = a1+a1+d = 2a1+d = 6+d
b2= b1q =q
b2+S2= q+6+d=12
q+d=6.(1)
q=S2/b2=(6+d)/q
q^2 = 6+d .(2)
由(1),(2)式解得：d=10或d=3
∵等比數(shù)列{bn}各項均為正數(shù)
∴q=3 ,d=3
an=a1+(n-1)d =3n （n∈N)
bn=b1q^(n-1) =3^(n-1) （n∈N)
2.∵Sn=na1 + n(n-1)d/2 = 3n(n+1)/2
∴1/Sn = 2/ 3n(n+1) =(2/3)[(1/n) - 1/(n+1)] （n∈N)
3.1/S1+1/S2+...+1/Sn = (2/3)(1 - 1/2) + (2/3)(1/2 - 1/3) + .+(2/3)[(1/n) - 1/(n+1)]
=(2/3)[1 - 1/2 + 1/2 - 1/3 +…+ 1/n - 1/(n+1)]
=(2/3)[1 - 1/(n+1)]
當(dāng)n=1時,1/S1+1/S2+...+1/Sn 取最小值 = 1/3
當(dāng)n=2,3,.n時,[1 - 1/(n+1)]的值無限接近于1
∴(1/3) ≤ 1/S1+1/S2+...+1/Sn ＜ (2/3)