Sum

Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120

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#### Solution

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

Sum = 20

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20

⇒ a = 5

Sum of the squares = 120

`(a – 3d)^2 + (a – d)^2 + (a + d)^2 + (a + 3d)^2 = 120`

`⇒ 4a^2 + d^2 = 120`

`⇒ a^2 + 5d^2 = 30`

⇒ 25 + 5d^{2} = 30 [∵ a = 5]

⇒ 5d^{2} = 5 ⇒ d = ± 1

If d = 1, then the numbers are 2, 4, 6, 8.

If d = – 1, then the numbers are 8, 6, 4, 2.

Thus, the numbers are 2, 4,6, 8 or 8, 6, 4, 2.

Concept: Sum of First n Terms of an AP

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