Find the sum of the 2 terms of an A P are in the ratio of (2n-1):(2n+1) then find the ratio of their 10th term ​

37/39

Step-by-step explanation:

We know that formula for n terms of an AP is written as;

(n/2)(2a + (n - 1)d)

Let the sum of the first AP be;

(n/2)(2a + (n - 1)d)

Let the sum of the second AP be (n/2)(2A + (n - 1)D)

We are told that they have a ratio of (2n-1):(2n+1)

Thus;

(2n-1)/(2n+1) = [(n/2)(2a + (n - 1)d)]/[ (n/2)(2A + (n - 1)D)]

On RHS, n/2 will cancel out to give;

(2n-1)/(2n+1) = [(2a + (n - 1)d)]/[(2A + (n - 1)D)]

Rewriting this we have;

(2n - 1)/(2n + 1) = [(a + ½(n - 1)d)]/[A + ½(n - 1)D)]

But we want tod find;

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

Thus,at n = 10, we have;

(a + 9d)/(A + 9D)

Comparing with [(a + ½(n - 1)d)]/[A + ½(n - 1)D)], we can say that;

½(n - 1) = 9

n = 18 + 1

n = 19

Thus;

(a + 9d)/(A + 9D) = (2n-1)/(2n+1)

Putting n = 19 gives;

(a + 9d)/(A + 9D) = (2(19) - 1)/(2(19) + 1)

(a + 9d)/(A + 9D) = 37/39

okay..i will pick 8!