Question:

Step 1: Use the G.P. condition for the three terms. Since $a+c$, $a+b$, $b+c$ are in G.P., the square of the middle term equals the product of the first and third terms: $$(a+b)^2 = (a+c)(b+c)$$ Step 2: Use the H.P. condition to express $c$ in terms of $a$ and $b$. Since $a$, $c$, $b$ are in H.P., the term $c$ is the harmonic mean of $a$ and $b$: $$c = \frac{2ab}{a+b}$$ Step 3: Expand the G.P. condition and simplify. Expanding the right side of the G.P. equation: $$(a+b)^2 = ab + ac + bc + c^2 = ab + c(a+b) + c^2$$ Let $s = a+b$ and $p = ab$. Then: $$s^2 = p + cs + c^2$$ Step 4: Substitute the H.P. relation into the simplified G.P. equation. From the H.P. condition, $c = \frac{2p}{s}$, which gives $p = \frac{cs}{2}$. Substituting into the equation from Step 3: $$s^2 = \frac{cs}{2} + cs + c^2 = \frac{3cs}{2} + c^2$$ Step 5: Rearrange and factor the equation. Rearranging: $$s^2 - c^2 = \frac{3cs}{2}$$ Factoring the left side: $$(s-c)(s+c) = \frac{3cs}{2}$$ Step 6: Introduce a variable for the desired ratio. Let $r = \frac{s}{c} = \frac{a+b}{c}$ (this is what we need to find). Then $s = rc$, and substituting: $$(rc - c)(rc + c) = \frac{3c \cdot rc}{2}$$ $$c^2(r-1)(r+1) = \frac{3rc^2}{2}$$ Step 7: Simplify and solve the quadratic equation. Dividing both sides by $c^2$: $$(r-1)(r+1) = \frac{3r}{2}$$ $$r^2 - 1 = \frac{3r}{2}$$ Multiplying by 2: $$2r^2 - 2 = 3r$$ $$2r^2 - 3r - 2 = 0$$ Step 8: Factor and find the solutions. Factoring the quadratic: $$(2r+1)(r-2) = 0$$ This gives $r = -\frac{1}{2}$ or $r = 2$. Step 9: Apply the constraint and state the final answer. Since $a, b, c > 0$, we must have $r = \frac{a+b}{c} > 0$. Therefore, $r = -\frac{1}{2}$ is rejected. $$\boxed{\frac{a+b}{c} = 2}$$ The answer is **Option 2: 2**