Question

The sum of four numbers which are in AP is 32 and the product of extreme is 55. Find the number

Answer 1

Answer :

Let, the numbers be (a - 3d), (a - d), (a + d) and (a + 3d)

Given that,

    The sum of the four numbers = 32

    ⇒ a - 3d + a - d + a + d + a + 3d = 32

    ⇒ 4a = 32

    ⇒ a = 8

Also given that,

    Product of the extremes = 55

    ⇒ (a - 3d) (a + 3d) = 55

    ⇒ a² - 9d² = 55

    ⇒ 8² - 9d² = 55     [ ∵ a = 8 ]

    ⇒ 9d² = 64 - 55

    ⇒ 9d² = 9

    ⇒ d² = 1

    ⇒ d = ± 1

When, a = 8 and d = 1, the four numbers be

  8 - 3 (1), 8 - 1, 8 + 1, 8 + 3 (1)

  i.e., 5, 7, 9 and 11

When a = 8 and d = - 1, the four numbers

  8 - 3 (- 1), 8 - (- 1), 8 + (- 1), 8 + 3 (- 1)

  i.e., 11, 9, 7 and 5

∴ the required four numbers are

    5, 7, 9 and 11

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