Find three consecutive whole numbers such that the sum of the squares of the numbers is equal to 869?

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Dr. Juanita Wolff · Accepted Answer

Numbers are: 16, 17. and 18

Dr. Juanita Wolff · Answer

Let the 3 consecutive #s be n-1 , n , n+1. Sum of squares = 3n^2 +2. Then 3n^2 = 867, ie., n^2 = 289 = (17)^2, ie., n = 17 & #s are 16,17,18.

Dr. Juanita Wolff · Answer

You can estimate it very quickly.Divide the sum by 3:869/3 = 289 2/3That must be close to the square of the middle number. Square root of that is 17.0195… The numbers must be 16 17 18.Work out he square to verify that.

Dr. Juanita Wolff · Answer

Let the consecutive whole number be ‘n’, ‘n+1’, ‘n+2’ .Their squares are n^2 , (n+1)^2 , (n+2)^2 Their sum is n^2 + (n^2 + 2n + 1) + ( n^2 + 4n + 4) = 869 3n^2 + 6n + 5 = 869 3n^2 + 6n = 864 n^2 + 2n = 288(n + 1)^2 – (1)^2 = 288(n + 1)^2 = 289n + 1 = sqrt(289) = 17 Hence n = 16 & n + 2 = 18 Hence the three consecutive numbers are 16,17,& 18.

Dr. Juanita Wolff · Answer

x² + (x + 1)² + (x + 2)² = 869x² + x² + 2x + 1 + x² + 4x + 4 = 8693x² + 6x – 864 = 0x² + 2x – 288 = 0[ x + 18 ] [ x – 16 ] = 0Accept x = 16Integers are 16 , 17 and 18

Dr. Juanita Wolff · Answer

Let n – 1, n, and n + 1 be the consecutive numbersThen (n – 1)^2 + n^2 + (n + 1)^2 = 869 n^2 – 2n + 1 + n^2 + n^2 + 2n + 1 = 869 3n^2 + 2 = 869 3n^2 = 867 n^2 = 287 n = √287 = 17 Numbers are: 16, 17. and 18

Dr. Juanita Wolff · Answer

(x – 1)^2 + x^2 + (x + 1)^2 = 869x^2 – 2x + 1 + x^2 + x^2 + 2x + 1 = 8693x^2 + 2 = 8693x^2 = 867x^2 = 289x = 17x – 1 = 16x + 1 = 18 hence the numbers are: 16, 17, 18

Dr. Juanita Wolff · Answer

Did you try 16, 17, and 18?You would be adding three numbers to add up to 869. So the average is 289.67. The square root of 289.67 is 17.02. So I would try numbers around 17. The first set to try would be 16, 17, and 18. This happens to work, so it’s the solution.

Dr. Katlynn Swaniawski II

Find three consecutive whole numbers such that the sum of the squares of the numbers is equal to 869?

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