Question

Three consecutive natural numbers are such that
the square of the middle number exceeds the
difference of the squares of the other two by
60. Find the number.​

Answer 1

$\bigstar\rm\blue{GIVEN}\bigstar$

• $\rm\blue{First\:Number=x}$

• $\rm\red{Second\:Number=(x+1)}$

• $\rm\green{Third\:Number=(x+2)}$

• The square of the middle exceeds the square of other two by 60.......1

$\bigstar\rm\blue{To\:Find}\bigstar$

• The three natural Number.

Atq.

$\implies\rm\red{((x+2)^2-(x))^2+60=(x+1)^2}$.......from1

$\implies\rm\red{x^2+4x+4-x^2+60=x^2+2x+1}$

$\implies\rm\red{2x+4+60=x^2+2x+1}$

$\implies\rm\red{x^2-2x-63=0}$

$\implies\rm\red{x^2-9x+7x-63=0}$

$\implies\rm\red{x(x-9)+7(x-9)=0}$

$\implies\rm\red{x+7=0,x-9=0}$

$\implies\rm\red{x=-7,x=9}$

As Natural Number can't be negative so,

x=9

Now,

First Natural Number =9

Second Natural Number= (9+1)=10

Third Natural Number= (9+2)=11

$\huge\rm\blue{VERIFICATION}$

$\implies\sf\blue{(Third\:Number)^2-(First\:Number)^2+60=(Second\:Number)^2}$

$\implies\sf\blue{(11)^2-(9)^2+60=(10)^2}$

$\implies\sf\blue{(121-81)+60=100}$

$\implies\sf\blue{40+60=100}$

$\implies\sf\blue{100=100}$

Hence, Verified✔️

Answer 2

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

According to the question ,

$\longrightarrow( x+1)^{2} =(x+2)^{2} -x^{2} +60$

$\longrightarrow x^{2} +1+2x = x^{2} +4+4x-x^{2} +60$

$\longrightarrow x^{2} +2x-4x+1-4-60 = 0$

$\longrightarrow x^{2} -2x-63 = 0$

$\longrightarrow x^{2} -9x+7x-63 = 0$

$\longrightarrow x(x-9)+7(x-9) = 0$

$\longrightarrow x-9 = 0, x + 7 = 0$

$\longrightarrow \underline{ \underline { x = -7 , x = 9 }}$

Natural number cannot be negative .

The natural numbers are 9 , 10  and 11 .