1.

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

Answer» Let the required numbers be
(a-3d) , (a-d) (a+d) and (a+3d)
Then ( a-3d) + (a -d) + ( a+d) + (a +3d) = 20` Rightarrow 4a = 20 Rightarrow a=5`
And ` (a-3d)^(2) + (a-d)^(2) + (a +d)^(2) + ( a+3d)^(2) =120 `
` Rightarrow 2(a^(2)+9a^(2)) =120`
` Rightarrow 4a^(2) + 20 d^(2) = 120 Rightarrow a^(2) + 5d^(2) = 30 `
` Rightarrow 25 + 5d^(2) =30Rightarrow 5d^(2) =5 Rightarrow d^(2) =1 Rightarrow d = +-1`
Hence, the required numbers are 2,4,6,8 or 8,6,4,2


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