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 September 1, 2021 by Nguyệt Ánh 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
Answer: 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 Reply
Answer:
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