Question

If three numbers are consecutive positive integers and 5 times the square of the largest number is greater than 2 times the sum of the squares of other two numbers by 75 , then find the sum of the largest and the smallest of these numbers.

Answer 1

Let,
first number be x
so,
second number is x+1 and
third number is x+2.

The question means=>
5(x+2)² - 2(x²+(x+1)²)=75

5[x²+4x+4]-2[x²+x²+2x+1]=75

​​​​​​(5x²+20x+20)-(4x²+4x+2)=75

x²+16x+18=75

x²+16x-57=0
=>
(x-3)(x+19)=0
{19× -3=-57 & 19+ -3=16}
=>
(x-3)=0 Or (x+19)=0
=>
x=3,-19
But -19 is negative integer,so we consider x=3;

x=3 which is the smallest number.
Largest number=x+2=5
There fore,
Sum of smallest number and largest number,
=3+5=8.