an=a1+(n-1)d
bn=b1q^(n-1)
a1=1=b1
a2=b2
=b1q
1+d=q (1)
a5=b3
a1+4d =b1q^2
1+4d= q^2 (2)
(1)^2/(2)
(1+d)^2 =1+4d
d^2-2d=0
d=2
q= 3
c1/b1+c2/b2+...+cn/bn= a(n+1) (1)
n=1
c1/b1 =a2
c1=3
c1/b1+c2/b2+...+c(n-1)/b(n-1)= an (2)
(1)-(2)
cn/bn = a(n+1)-an
= 2
cn =2bn
=2.3^(n-1)
ie
cn = 3 ; n=1
=2.3^(n-1) ; n=2,3,4,...
c1+c2+...+cn
=3+c2+...+cn
= 3 + 3(3^(n-1) -1)
=3^n