Question

3 consecutive integers are such that the square of first number and product of the remaining two add up to 121. find these numbers

Answer 1

Let's take the first integer as x

So the other two consecutive integer would be x+1 and x+2 :

According to question:

x^2 + { (x+1)×(x+2) } = 121

x^2 + x^2 + 2x + x + 2 = 121

2x^2 + 3x = 121-2 = 119

Simplifying it ( taking 2 as common )

2( x^2+1.5x ) = 119

By transposition :

x^2+1.5x = 119/2 = 59.5

If we observe the squares of no.s from 1 to 10 .

We can conclude that only the value of square of 7 comes close to the value but doesn't surpass it .

So taking x as 7 for trial

7^2 + 1.5×7 = 59.5

49+10.5 = 59.5

59.5 = 59.5

Hence the value of x is satisfied by 7.

Now x+1 = 7+1 = 8

and x+2 = 7+2 = 9