Question

Sol. [Key Concept: Set $S_{30} = (T_{30})^{3}$ with first term $10/3$ and solve for the common difference.]

$$\ a = \frac{10}{3}:S_{30} = \left( T_{30} \right)^{3}$$

$$\begin{matrix} & \frac{30}{2}\left\lbrack 2 \times \frac{10}{3} + 29\text{ }d \right\rbrack = \left\lbrack \frac{10}{3} + 29\text{ }d \right\rbrack^{3} \\ & 15\left( \frac{20}{3} + 29\text{ }d \right) = \left( \frac{10}{3} + 29\text{ }d \right)^{3} \\ & 15\left( \frac{10}{3} + \frac{10}{3} + 29\text{ }d \right) = \left( \frac{10}{3} + 29\text{ }d \right)^{3} \end{matrix}$$

By observation $\frac{10}{3} + 29\text{ }d = 5$

$${29\text{ }d = 5 - \frac{10}{3} = \frac{5}{3} }{d = \frac{5}{87} }$$