If the sums of n terms of 2 A.P. are in ratio (3n+5)/(5n+7), then find their nth terms ratio

To find the ratio of the nth terms of two arithmetic progressions (APs) when the sums of their terms are in a given ratio, we can use the formula for the sum of an AP.

Let's assume the first arithmetic progression (AP) has a common difference of 'd' and the second AP has a common difference of 'D'.

The sum of the first AP up to the nth term (Sn1) is given by:
Sn1 = (n/2) * [2a + (n-1)d], where 'a' is the first term of the AP.

The sum of the second AP up to the nth term (Sn2) is given by:
Sn2 = (n/2) * [2A + (n-1)D], where 'A' is the first term of the second AP.

Given that Sn1/Sn2 = (3n+5)/(5n+7), we can equate the two expressions and solve for d/D:

[(n/2) * [2a + (n-1)d]] / [(n/2) * [2A + (n-1)D]] = (3n+5)/(5n+7)

After simplifying the equation, we get:
(2a + (n-1)d) / (2A + (n-1)D) = (3n+5)/(5n+7)

Now, we can find the ratio of the nth terms by substituting the value of 'n' into the equation and simplifying:

(a + (n-1)d) / (A + (n-1)D)

Therefore, the ratio of the nth terms of the two arithmetic progressions is (a + (n-1)d) / (A + (n-1)D).