Question:

Step 1: Given that $ab^2c^3$, $a^2b^3c^4$, and $a^3b^4c^5$ are in Arithmetic Progression (A.P.), the middle term is the average of the other two terms. Thus, we have: $$2(a^2b^3c^4) = ab^2c^3 + a^3b^4c^5$$ Step 2: Since $a, b, c > 0$, we can divide the entire equation by $ab^2c^3$: $$2 \frac{a^2b^3c^4}{ab^2c^3} = \frac{ab^2c^3}{ab^2c^3} + \frac{a^3b^4c^5}{ab^2c^3}$$ $$2abc = 1 + a^2b^2c^2$$ Step 3: Rearrange the equation to form a quadratic in $abc$: $$(abc)^2 - 2(abc) + 1 = 0$$ This is a perfect square: $$(abc - 1)^2 = 0$$ Therefore, $$abc = 1$$ Step 4: To find the minimum value of $a+b+c$, we apply the AM-GM inequality. For positive numbers $a, b, c$: $$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$ Substituting $abc = 1$: $$\frac{a+b+c}{3} \ge \sqrt[3]{1}$$ $$\frac{a+b+c}{3} \ge 1$$ $$a+b+c \ge 3$$ Step 5: The equality in the AM-GM inequality holds when $a=b=c$. If $a=b=c$, then from $abc=1$, we have $a \cdot a \cdot a = 1$, which implies $a^3=1$. Since $a>0$, we get $a=1$. Thus, $a=b=c=1$. In this case, $a+b+c = 1+1+1 = 3$. The minimum value of $a+b+c$ is 3.