Question

Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is grater than the sum of the squares of the other two numbers by 67; calculate the value of p.

Answer 1

Given Let $p-1,p,p+1$ be three consequetive position integer.

According to question

$3(p+1{)}^2=(p-1{)}^2+p^2+67$

$3[p^2+1+2p]=p^2+1-2p+p^2+67$

$3p^2+6p+3=2p^2-2p+68$

$p^2+8p-65=0$

$p(p+13)-5(p+13)=0$

$(p-5)(p+13)=0$

p=5 & p=-13

We can't take p=-133 as number are positive then [p=5]

now numbers are 5-1,5,5+1

or [4,5 & 6]