b 1=a 1= 1

b2=q* 1=a3= 1+2d

b4=q^3* 1=a27= 1+26d

q^3=( 1+2d)^3= 1+26d

Simplified to d*(2d+5)(d- 1)=0, because d

An=n,Bn=3^(n- 1)

Sm = ∑ (1/m) = +∞, TN = ∑ 2/3 (n- 1) = 3, so there is m= 19 (estimated, the actual value is within +-2, in short, it exists).

2.

Is this the meaning of the title: (6An-3)*Bn/(An*A(n+ 1))?

pn = 2(6n-3)*3^(n- 1)/n(n+ 1)= 2(2n- 1)*3^n/n(n+ 1)

Take the logarithm, and estimate and solve it to get n=9.