Question

If the sum of four integers in A.P is 48 and their product is 15120, then find the numbers
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Answer 1

Answer:

Sum of an AP upto fourth term is given as 48.

Let the terms be a -3d, a-d , a+d, a+3d

So the sum of these terms is a -3d + a - d + a + d + a + 3d = 48

So 4a = 48

a = 48/4 = 12

So (12 - 3d) ⨯ (12 - d) ⨯(12 + d)⨯ (12 +3d) = 15120

And (a - b)(a + b) = (a2 - b2 )

so (144 - 9d2)(144 - d2) = 15120

Let d2 be x, so the equation get changed (144 - 9x) (144 - x) = 15120

⇒ 20736 - 144x - 1296x +9 x2 = 15120

⇒ 20736 - 1440x + 9 x2 = 15120

⇒5616 - 1440x + 9x2 = 0

So solving it we get x = 4, x = 156

Hence d2 = 4, or d = 2,-2 we get the numbers as 6,10,14,18 or 18,14,10,6

And d2 = 156 or d = 12.49,-12.49 , we get the numbers as -25.47, -0.49, 24.49 , 49.47 and 49.47 , 24.49, -0.49, -25.47