1. Interpret the distances

Given:

• The object moves along segments $AB$, $BC$, and $CD$ with speeds $v_1, v_2, v_3$ respectively.
• $AB = BC$
• $AD = 3AB$

Since $AD = AB + BC + CD$ and $AB=BC$, let $AB = BC = x$ Then $AD = x + x + CD = 3x$ So, $CD = x$ Hence all three segments are equal: $AB = BC = CD = x$

2. Find total distance

Total distance travelled is $AB+BC+CD = x+x+x = 3x$

3. Find total time

Time taken on each segment: $t_1 = \frac{x}{v_1}, \qquad t_2 = \frac{x}{v_2}, \qquad t_3 = \frac{x}{v_3}$

Therefore total time is $T = \frac{x}{v_1}+\frac{x}{v_2}+\frac{x}{v_3}$ $T = x\left(\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}\right)$

4. Average speed formula

Average speed is $v_{\text{avg}} = \frac{\text{total distance}}{\text{total time}}$

So, $v_{\text{avg}} = \frac{3x}{x\left(\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}\right)}$

Cancel $x$: $v_{\text{avg}} = \frac{3}{\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}}$

Now simplify: $\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3} = \frac{v_2v_3+v_3v_1+v_1v_2}{v_1v_2v_3}$

Therefore, $v_{\text{avg}} = 3\cdot \frac{v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1}$

So, $\boxed{v_{\text{avg}} = \frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1}}$

5. Match with options

This matches Option D: $\boxed{\frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1}}$

6. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.