the sum of the squares of 3 consecutive positive numbers is 365 find all those numbers
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let the three numbers be (x-1), x ,( x+1)
therefore, (x-1)² +x² + (x+1)² = 365
3x²+2 = 365
3x² =363
x² = 121
x= 11
x=-11 is neglected, as x is positive
therefore the numbers are 10,11,12
therefore, (x-1)² +x² + (x+1)² = 365
3x²+2 = 365
3x² =363
x² = 121
x= 11
x=-11 is neglected, as x is positive
therefore the numbers are 10,11,12
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