设{an}中 :an = a1 +(n-1)d;则bn = (1/2)^an;b1+b2+b3 = (1/2)^a1[1 + (1/2)^d + (1/2)^(2d)] = 21/8;b1b2b3 = (1/8)^(a1 + d) = 1/8;==>d = 2,a1 = -1;==>an = -1 + 2(n-1) = 2n -3