The sum of squares of three consecutive natur | vedclass

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(B) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the sum of their squares is $2030$:$(x-1)^2 + x^2 +

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the sum of their squares is $2030$:$(x-1)^2 + x^2 + (x+1)^2 = 2030$Expanding the squares:$(x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 2030$Combining like terms:$3x^2 + 2 = 2030$Subtracting $2$ from both sides:$3x^2 = 2028$Dividing by $3$:$x^2 = 676$Taking the square root:$x = \sqrt{676} = 26$Therefore,the middle number is $26$.

The sum of squares of three consecutive natural numbers is $2030$. What is the middle number?

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