Question

the product of 3 consecutive integers is 120 then the greatest number is ?

A) 4
B) 6
C) 8
D) 5​

Answer 1

Answer:

Given:-

The product of three consecutive integer is 120

To Find :-

The greatest number .

Solution:-

Now, we know that,

As they are consecutive so, the integer would be one after the other .

Let, the integer be x

The second integer would be ( x + 1 )

The third integer would be ( x + 2 )

Now,

Their product is 120 ( given )

ATQ

⟹ x × ( x + 1 ) × ( x + 2 ) = 120

⟹ x ² + x ( x + 2 ) = 120

⟹ x² ( x + 2 ) + x ( x + 2 ) = 120

⟹ x³ + 2x² + x² + 2x = 120

⟹ x² ( x + 2 ) + x ( x + 2 ) = 120

⟹ ( x + 2 ) ( x² + x ) = 120

now,

x + 2 = 120

x = 118

Putting the value of x in ( x² + x ) = 120 , we get

x² + x = 120

x² + x - 120 = 0 .

x² + 118 - 120 = 0

x² - 2 = 0

x² = 2

x = $\sqrt{ 2 }$

so,

putting the value of x in x , ( x + 1 ) , ( x +2 )

x = $\sqrt{ 2 }$

x + 1 = $\sqrt{2 }$ + 1

x + 2 = $\sqrt{2 }$ + 2