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

the answer is. -82562234434584

step-by-step explanation:

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.