Question

the sum of first n terms of two AP are in the ratio (7n-5):(5n+7). show that the 6n termof two APs are equal ​

Answer 1

Answer:

Step-by-step explanation:

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 = 5

6th term of 1st AP =a+5d=1+5×7=36

6th term of 2nd AP=A+5D=11+5×5=36

Answer 2

Answer:

36

Step-by-step explanation:

6th term of arithmetic progression is 36