Find Common Difference of Arithmetic Progression

To find the common difference of the Arithmetic Progression (A.P.), let us define the given parameters:

• Number of terms, $n = 30$
• First term, $a = \frac{10}{3}$
• Let $d$ be the common difference.
• Let $L$ be the last term ($30^{\text{th}}$ term) of the A.P.

The $30^{\text{th}}$ term $L$ can be expressed as: $L = T_{30} = a + (30 - 1)d = a + 29d$

The sum of the first $30$ terms of an A.P. is given by: $S_{30} = \frac{n}{2} (a + L) = \frac{30}{2} (a + L) = 15(a + L)$

According to the given condition, the sum of this A.P. is equal to the cube of its last term: $S_{30} = L^3$

Substituting the expression for $S_{30}$: $15(a + L) = L^3$

Substitute $a = \frac{10}{3}$ into the equation: $15\left(\frac{10}{3} + L\right) = L^3$ $50 + 15L = L^3$ $L^3 - 15L - 50 = 0$

We now solve this cubic equation for $L$. Testing for rational roots, we find that $L = 5$ satisfies the equation: $5^3 - 15(5) - 50 = 125 - 75 - 50 = 0$

Thus, $(L - 5)$ is a factor. Factoring the cubic polynomial gives: $(L - 5)(L^2 + 5L + 10) = 0$

For the quadratic factor $L^2 + 5L + 10 = 0$, the discriminant is: $\Delta = 5^2 - 4(1)(10) = 25 - 40 = -15 < 0$

Since the discriminant is negative, there are no other real solutions for $L$. Thus, the only real solution is: $L = 5$

Now, using the relation $L = a + 29d$: $5 = \frac{10}{3} + 29d$ $29d = 5 - \frac{10}{3}$ $29d = \frac{5}{3}$ $d = \frac{5}{3 \times 29} = \frac{5}{87}$

Since $a = \frac{10}{3} > 0$ and $d = \frac{5}{87} > 0$, all terms of the A.P. are non-negative, satisfying the problem statement.

Therefore, the correct option is A.