Three charges +q, +2q, and +4q are placed in a line separated by distance d. Calculate the tensions in the strings AB and BC.

Consider the forces acting on charge C ($+4q$). It is repelled by charge B ($+2q$) and charge A ($+q$). The string BC pulls charge C to the left.
The distance between B and C is $d$.
The distance between A and C is $d+d = 2d$.
The electrostatic force of repulsion between B and C is $F_{BC} = \frac{1}{4\pi\epsilon_0} \frac{(2q)(4q)}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{8q^2}{d^2}$. This force acts to the right on C.
The electrostatic force of repulsion between A and C is $F_{AC} = \frac{1}{4\pi\epsilon_0} \frac{(q)(4q)}{(2d)^2} = \frac{1}{4\pi\epsilon_0} \frac{4q^2}{4d^2} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2}$. This force also acts to the right on C.
For charge C to be in equilibrium, the tension $T_{BC}$ must balance these repulsive forces:

$$T_{BC} = F_{BC} + F_{AC} = \frac{1}{4\pi\epsilon_0} \frac{8q^2}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{9q^2}{d^2}$$

Consider the forces acting on charge B ($+2q$). It is repelled by charge A ($+q$) to the right and by charge C ($+4q$) to the left. The string AB pulls charge B to the left, and the string BC pulls charge B to the right (this is the tension $T_{BC}$ acting on B).
The electrostatic force of repulsion between A and B is $F_{AB} = \frac{1}{4\pi\epsilon_0} \frac{(q)(2q)}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{2q^2}{d^2}$. This force acts to the right on B.
The electrostatic force of repulsion between B and C is $F_{BC}' = \frac{1}{4\pi\epsilon_0} \frac{(2q)(4q)}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{8q^2}{d^2}$. This force acts to the left on B.
For charge B to be in equilibrium, the sum of forces to the right must equal the sum of forces to the left:

$$F_{AB} + T_{BC} = T_{AB} + F_{BC}'$$

Substitute the values:

$$\frac{1}{4\pi\epsilon_0} \frac{2q^2}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{9q^2}{d^2} = T_{AB} + \frac{1}{4\pi\epsilon_0} \frac{8q^2}{d^2}$$ $$\frac{1}{4\pi\epsilon_0} \frac{11q^2}{d^2} = T_{AB} + \frac{1}{4\pi\epsilon_0} \frac{8q^2}{d^2}$$ $$T_{AB} = \frac{1}{4\pi\epsilon_0} \frac{11q^2}{d^2} - \frac{1}{4\pi\epsilon_0} \frac{8q^2}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{3q^2}{d^2}$$

$$\frac{T_{AB}}{T_{BC}} = \frac{\frac{1}{4\pi\epsilon_0} \frac{3q^2}{d^2}}{\frac{1}{4\pi\epsilon_0} \frac{9q^2}{d^2}} = \frac{3}{9} = \frac{1}{3}$$

So, the ratio $T_{AB} : T_{BC}$ is $1:3$.