The sum of the squares of three consecutive natural numbers is 2030, then the middle number is
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Hey,sup!
As per the question,
Let the consecutive numbers be x,x+1,x+2.
So, x^2+(x+1)^2+(x+2)^2=2030.
=> x^2+x^2+1+2x+x^2+4+4x=2030.
=> 3x^2+5+6x-2030=0.
=> 3x^2 +6x -2025 = 0.
Dividing both side by 3
=>x^2 +2x - 675=0.
=>x^2 + 27x -25x - 675 = 0.
=> x(x+27) - 25(x+27)=0.
=> (x-25) (x+27) =0.
We'll discard x+27 as it gives negative value.
=> x-25=0.
=> x=25.
So the consecutive numbers are 25,26,27.
The middle number is 26.
Hope it helps.
As per the question,
Let the consecutive numbers be x,x+1,x+2.
So, x^2+(x+1)^2+(x+2)^2=2030.
=> x^2+x^2+1+2x+x^2+4+4x=2030.
=> 3x^2+5+6x-2030=0.
=> 3x^2 +6x -2025 = 0.
Dividing both side by 3
=>x^2 +2x - 675=0.
=>x^2 + 27x -25x - 675 = 0.
=> x(x+27) - 25(x+27)=0.
=> (x-25) (x+27) =0.
We'll discard x+27 as it gives negative value.
=> x-25=0.
=> x=25.
So the consecutive numbers are 25,26,27.
The middle number is 26.
Hope it helps.
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2
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26
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