the sum of 4 numbers in a.p is 40 and the product of their extremes is 91.the number are

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

Sum = (a - 3d) + (a - d) + (a + d) + (a + 3d) = 40

⇒ 4a = 40

⇒ a = 10

Given, product of extremes = 91

⇒ (a - 3d)(a + 3d) = 91

⇒ a^{2} - 9d^{2} = 91

⇒ (10)^{2} - 9d^{2} = 91

⇒ - 9d^{2} = 91 -100 = -9

⇒ d^{2} = 1

⇒ d = ±1

**Case1:** When a = 10 and d = 1, then numbers are:

10 - 3, 10 - 1, 10 + 1, 10 + 3

or 7,9,11,13

**Case2:** When a = 10 and d = -1, then numbers are:

10 + 3, 10 + 1, 10 - 1, 10 - 3

or 13, 11, 9, 7

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