S1 is a series of 4 consecutive even numbers. If the sum of the reciprocals of the first two numbers of S1 is $\frac{11}{60}$, then what is the reciprocal of the third highest number of S1?

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**Solution:**

1. Let the 4 consecutive even numbers be $x$, $x+2$, $x+4$, and $x+6$.

2. Given that the sum of the reciprocals of the first two numbers is:
   $$\frac{1}{x} + \frac{1}{x+2} = \frac{11}{60}$$

3. Find a common denominator and combine:
   $$\frac{(x+2) + x}{x(x+2)} = \frac{11}{60}$$
   $$\frac{2x + 2}{x^2 + 2x} = \frac{11}{60}$$

4. Multiply both sides by $60(x^2 + 2x)$:
   $$60(2x + 2) = 11(x^2 + 2x)$$
   $$120x + 120 = 11x^2 + 22x$$

5. Rearranging to form a quadratic equation:
   $$11x^2 + 22x - 120x - 120 = 0$$
   $$11x^2 - 98x - 120 = 0$$

6. Solve the quadratic equation:
   Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=11$, $b=-98$, $c=-120$:

$$x = \frac{98 \pm \sqrt{(-98)^2 - 4 	imes 11 	imes (-120)}}{2 	imes 11}$$
   $$= \frac{98 \pm \sqrt{9604 + 5280}}{22}$$
   $$= \frac{98 \pm \sqrt{14884}}{22}$$

Since $\sqrt{14884} = 122$ (approximately),

$$x = \frac{98 \pm 122}{22}$$

- Taking positive root:
     $$x = \frac{98 + 122}{22} = \frac{220}{22} = 10$$

- Taking negative root:
     $$x = \frac{98 - 122}{22} = \frac{-24}{22} = -\frac{12}{11}$$ (not an even integer, discard)

7. So, $x = 10$ is the first even number.

8. The four numbers are: 10, 12, 14, 16.

9. The third highest number is the third number in the series, which is 14.

10. The reciprocal of 14 is:
    $$\frac{1}{14}$$

**Answer:** $\frac{1}{14}$

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