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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Nguyệt Ánh 5 months 2021-09-01T03:56:42+00:00 1 Answers 0 views 0

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    2021-09-01T03:58:29+00:00

    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

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )