Step 1: Understand the chain.
Three identical rods A, B, C are joined end to end with conductivities $k$, $2k$, and $k/2$. End A is at 100 degrees and end C is at 0 degrees. We find the temperature at the A B junction in steady state.
Step 2: Use steady heat flow.
In steady state the same heat current $H$ flows through all three rods. For each rod $H = \dfrac{k A \Delta T}{L}$, and since the rods are identical in $A$ and $L$, only the conductivity and the temperature drop matter.
Step 3: Name the junctions.
Let $T_1$ be the temperature at the A B junction and $T_2$ at the B C junction. Then $H = k(100 - T_1) = 2k(T_1 - T_2) = \tfrac{k}{2}(T_2 - 0)$.
Step 4: Relate T2 to T1.
From $k(100 - T_1) = \tfrac{k}{2} T_2$ we get $T_2 = 2(100 - T_1) = 200 - 2T_1$.
Step 5: Use the middle rod equation.
From $2k(T_1 - T_2) = \tfrac{k}{2} T_2$, multiply out: $4(T_1 - T_2) = T_2$, so $4 T_1 = 5 T_2$.
Step 6: Solve for T1.
Substitute $T_2 = 200 - 2T_1$: $4 T_1 = 5(200 - 2T_1) = 1000 - 10 T_1$. So $14 T_1 = 1000$, giving $T_1 \approx 71^\circ$C. \[ \boxed{71^\circ \text{C}} \]