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The sum of the squares of three consecutive natural numbers is 149. Find the numbers.
Given:
The sum of the squares of three consecutive natural numbers is 149.
To do:
We have to find the numbers.
Solution:
Let the three consecutive natural numbers be $x-1$, $x$ and $x+1$.
According to the question,
$(x-1)^2+(x)^2+(x+1)^2=149$
$x^2-2x+1+x^2+x^2+2x+1=149$
$3x^2+2-149=0$
$3x^2-147=0$
$3(x^2-49)=0$
$x^2-49=0$
$(x)^2-(7)^2=0$
$(x+7)(x-7)=0$
$x+7=0$ or $x-7=0$
$x=-7$ or $x=7$
$-7$ is not a natural number. Therefore, $x=7$.
$x-1=7-1=6$ and $x+1=7+1=8$
The required natural numbers are $6$, $7$ and $8$.
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