Three consecutive positive integers k, l and m are such that l\(^2\) = 3(k+m). Find the value of m.

  • A 4
  • B 5
  • C 6
  • D 7

The correct answer is D. 7

Let's say the three consecutive positive integers are k, k+1, and k+2. Since l is the middle integer, we can say that l = k+1. Similarly, m is the largest integer, so m = k+2. Substituting these values into the given equation l\(^2\) = 3(k+m), we get:

(k+1)\(^2\) = 3(k + (k+2))

Expanding and simplifying this equation, we get:

k\(^2\) + 2k + 1 = 3(2k + 2)

k\(^2\) + 2k + 1 = 6k + 6

k\(^2\) - 4k - 5 = 0

This is a quadratic equation that can be solved by factoring or using the quadratic formula. Factoring, we get:

(k - 5)(k + 1) = 0

So, k = 5 or k = -1.

Since k is a positive integer, we can discard the negative solution and conclude that k = 5.

Substituting this value into the expression for m, we get:

m = k + 2

m = 5 + 2

m = 7

Therefore, the correct option is D. 7.

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