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

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Solution

Let the numbers be p – 1, p and p + 1.
From the given information,
$3 (p + 1)^{2} = (p - 1)^{2} + p^{2} + 67$
$3 p^{2} + 6 p + 3 = p^{2} + 1 - 2 p + p^{2} + 67$
$p^{2} + 8 p - 65 = 0$
$(p + 13) (p - 5) = 0$
$p = - 13, 5$
Since, the numbers are positive so $p$ cannot be equal to $- 13$ .
Thus, $p = 5$ .

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

Let the numbers be p – 1, p and p + 1. From the given information, 3(p+1)2=(p–1)2+p2+67 3p2+6p+3=p2+1–2p+p2+67 p2+8p–65=0 (p+13)(p–5)=0 p=−13,5 Since, the numbers are positive so p cannot be equal to −13. Thus, p=5.