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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
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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
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Mathematics
5 months
2021-09-01T03:56:42+00:00
2021-09-01T03:56:42+00:00 1 Answers
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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