The sum of n terms of two AP are in the ratio (7n–5):(5n–17) show that their term are equal.

The sum of n terms of two AP are in the ratio (7n–5):(5n–17) show

1.	The sum of n terms of two AP are in the ratio (7n–5):(5n–17) show that their term are equal.
Answer» Tum mera name kaise jante ho Tum dono mai se Lucifer kon hai Nothing Let a , d and A , D be the first term and the common ratio of the first and the second AP respectively.So [(n/2){2a+(n-1)d}]/[(n/2){2A+(n-1)D}]=(7n-5)/(5n+17)={7(n-1)+2}/{5(n-1)+22}So {2a+(n-1)d}/{2A+(n-1)D}={2+7(n-1)}/{22+5(n-1)} =>a=1, d=7, A =11, D = 56th term of 1st AP =a+5d=1+5×7=366th term of 2nd AP=A+5D=11+5×5=36 U didn\'t answer me what happened