Question

the sum of the squares of three consecutive natural number is 110 find the numbers

Answer 1

$\bf\huge{ANSWER}$

Let the three consecutive numbers be x , ( x + 1 ) and ( x + 2 ) .

$\rm\longrightarrow x^2+(x+1)^2+(x+2)^2=110\\\\\longrightarrow x^2+x^2+2x+1+x^2+4x+4=110\\\\\longrightarrow 3x^2+6x+5=110\\\\\longrightarrow 3x^2+6x=110-5\\\\\longrightarrow 3x^2+6x=105\\\\\longrightarrow 3x^2+6x-105=0\\\\\longrightarrow x^2+2x-35=0\\\\\longrightarrow x^2+7x-5x-35=0\\\\\longrightarrow x(x+7)-5(x-7)=0\\\\\longrightarrow (x+7)(x-5)=0\\\\\longrightarrow x=-7 \ or \ x=5$

Natural numbers cannot be negative .

Required numbers are 5 , 6 and 7 .

Answer 2

Given,

• Sum of the squares of three consecutive natural number = 110

Let the three consecutive numbers be x , ( x + 1 ) and ( x + 2 ) .

$\implies\displaystyle\sf{x^2+(x+1)^2+(x+2)^2=110}$

$\implies\displaystyle\sf{(x^2)+(x^2+2x+1)+(x^2+4x+4)=110}$

$\implies\displaystyle\sf{x^2+x^2+2x+1+x^2+4x+4=110}$

$\implies\displaystyle\sf{3x^2+6x+5=110}$

$\implies\displaystyle\sf{3x^2+6x+5-110=0}$

$\implies\displaystyle\sf{3x^2+6x-105=0}$

Dividing all the terms by 3 ,

$\implies\displaystyle\sf{x^2+2x-35=0}$

$\implies\displaystyle\sf{(x+7)(x-5)=0}$

$\implies\displaystyle\sf{x=-7\:or\:5}$

The numbers cannot be negative since they are natural numbers,

$\implies\displaystyle\sf{x=5}$

Therefore the terms are :

$\implies\displaystyle\sf{x=5}$

$\implies\displaystyle\sf{x+1=6}$

$\implies\displaystyle\sf{x+3=7}$

$\boxed{\displaystyle\sf{\therefore\:The\:Numbers\:are\:\:5\:,6\:,\:7}}$