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Question
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120
Sum
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 + 5d2 = 30 [∵ a = 5]
⇒ 5d2 = 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.
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