Question

The middle number, out of three consecutive positive integers is $p$ . Three times the square of the largest number is greater than the sum of the squares of the other two numbers by $67$ Calculate the value of $p$ .

Answer 1

The series of numbers that are more than one from its preceding number is known as consecutive numbers. The consecutive number series can be even as well as odd depending on the first term of the series and the difference between the numbers. If the difference between the two consecutive numbers is $2$ and the starting term is an odd integer then, it is known as the series of the odd consecutive numbers, whereas if the difference between the two consecutive numbers is $2$ and the starting term is an even integer, then it is known to be an even consecutive number series.

Complete step-by-step solution:
Let the three consecutive positive integers be $p - 1,p,p + 1$ .
We are given that $p$ is the middle term.
The largest term is $p + 1$ and the smallest term is $p - 1$
Now, according to the question
$\Rightarrow 3{\left( {p + 1} \right)^2} = {\left( {p - 1} \right)^2} + {p^2} + 67$ -----(i)
We know the following algebraic properties:
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Using the same in (i) we get,
$\Rightarrow 3{\left( {p + 1} \right)^2} = {\left( {p - 1} \right)^2} + {p^2} + 67 \\\ \Rightarrow 3\left( {{p^2} + 1 + 2p} \right) = {p^2} + 1 - 2p + {p^2} + 67 \\\ \Rightarrow 3{p^2} + 3 + 6p = 2{p^2} - 2p + 68 \\\ \Rightarrow 3{p^2} - 2{p^2} + 6p + 2p + 3 - 68 = 0 \\\ \Rightarrow {p^2} + 8p - 65 = 0 \\\$
As now, we get a quadratic equation. We will solve it using the method of splitting the middle term.
$\Rightarrow {p^2} + 13p - 5p - 65 = 0 \\\ \Rightarrow p\left( {p + 13} \right) - 5\left( {p + 13} \right) = 0 \\\ \Rightarrow \left( {p - 5} \right)\left( {p + 13} \right) = 0 \\\$
So, we get $p = 5$ and $p = - 13$ .
We can’t take $p = - 13$ as we are given that the numbers are positive integers.
So, $p = 5$ and the consecutive positive integers will be
$\Rightarrow p - 1 = 5 - 1 = 4 \\\ \Rightarrow p = 5 \\\ \Rightarrow p + 1 = 5 + 1 = 6 \\\$

Hence, the consecutive positive integers are $4,5$ and $6$ .

Note: It is interesting to note here that for a series of even or odd consecutive numbers, all the terms present in the series are even and odd respectively.
$1,2,3,4,...$ is a series of general consecutive numbers.
$2,4,6,8,...$ is a series of even consecutive numbers.
$1,3,5,7,...$ is a series of odd consecutive numbers.