Let us take two arithmetic sequence :
#(i)a, a+d,a+2d,a+3d,...a+(n-1)d,...#
#(ii)A,A+D,A+2D,A+3D,...,A+(n-1)D,...#
So,the #n^(th)#term :
#a_n=a+(n-1)d and A_n=A+(n-1)D#
#:.ratio=(a+(n-1)d)/(A+(n-1)D)=(2n+8)/(5n-3)#
#=>(2a+(2n-2)d)/(2A+(2n-2)D)=(2n+8)/(5n-3)...to(1)#
Now, take new term #=N^(th)term#
So, the sum of #N^(th)term# :
#S_N=N/2[2a+(N-1)d] and S'_N=N/2{2A+(N-1)D]#
#:. ratio =S_N/(S'_N)=(N/2[2a+(N-1)d])/(N/2{2A+(N-1)D])#
#S_N/(S'_N)=(2a+(N-1)d)/(2A+(N-1)D)...to(2)#
Comparing LHS of #(1) and# RHS of #(2)# we get
#2n-2=N-1=>2n=N-1+2#
#=>2n=N+1=>n=(N+1)/2#
Subst. #n=(N+1)/2# into #(1)#
#(2a+(2((N+1)/2)-2)d)/(2A+(2((N+1)/2)-2)D)=
(2((N+1)/2)+8)/(5((N+1)/2)-3)#
#=>(2a+(N-1)d)/(2A+(N-1)D)=(2N+2+16)/(5N+5-6)#
From #(2)#
#S_N/(S'_N)=(2N+18)/(5N-1)#
