80n + \frac{n(n-1)}{2} = 120 \cdot 60

To solve this equation, we first simplify the left side:

80n + \frac{n(n-1)}{2} = 120 \cdot 60 80n + \frac{n^2 - n}{2} = 7200 Multiplying both sides by 2 to get rid of the fraction: 160n + n^2 - n = 14400 Rearranging the terms: n^2 + 159n - 14400 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:

n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a = 1, b = 159, and c = -14400. Plugging these values into the formula:

n = \frac{-159 \pm \sqrt{159^2 - 4(1)(-14400)}}{2(1)} n = \frac{-159 \pm \sqrt{25281 + 57600}}{2} n = \frac{-159 \pm \sqrt{82881}}{2} n = \frac{-159 \pm 287}{2}

Therefore, the solutions for n are: n = \frac{-159 + 287}{2} = \frac{128}{2} = 64 n = \frac{-159 - 287}{2} = \frac{-446}{2} = -223

Since the number of items cannot be negative, the only valid solution is n = 64.