Let the first term of first AP series be #a# and its common difference be #b#.
And let the first term of 2nd AP series be #c# and its common difference be #d#.
If the sum of n terms of these two series be #s_n andS_n# respectively then
#s_n/(S_n)=(n/2(2a+(n-1)b))/(n/2(2c+(n-1)*d))=(7n+1)/(4n+27)....(1)#
Again if the n th terms of these two series be #t_n andT_n# respectively then
#t_n/T_n=(a+(n-1)b)/(c+(n-1)*d)#
#=>t_n/T_n=(n/2(2a+(2n-2)b))/(n/2(2c+(2n-2)*d))#
#=>t_n/T_n=(n/2(2a+((2n-1)-1)b))/(n/2(2c+((2n-1)-1)d))#
#=>t_n/T_n=s_(2n-1)/S_(2n-1)#
#=>t_n/T_n=(7(2n-1)+1)/(4(2n-1)+27)#[Putting #(2n-1)# in place of n in (1)]
#=>t_n/T_n=(14n-6)/(8n+23)#
Putting #n=m# we get the desired ratio
#t_m/T_m=(14m-6)/(8m+23)#
