The first term of an A.P. of 30 non-negative terms is $$\frac{10}{3}$$. If the sum of the A.P. is the cube of its last term, then its common difference is :

For an arithmetic progression (A.P.) let

first term $$a=\frac{10}{3},$$
common difference $$d,$$
number of terms $$n=30.$$

The last (30th) term is

$$l=a+(n-1)d=\frac{10}{3}+29d.$$

Sum of an A.P. with $$n$$ terms is $$S=\frac{n}{2}(a+l).$$ Hence

$$S=\frac{30}{2}\left(\frac{10}{3}+l\right)=15\left(\frac{10}{3}+l\right).$$

Because $$l=\frac{10}{3}+29d,$$ the bracket becomes

$$\frac{10}{3}+l=\frac{10}{3}+\left(\frac{10}{3}+29d\right)=\frac{20}{3}+29d.$$

Therefore

$$S=15\left(\frac{20}{3}+29d\right)=100+435d.$$

The question states that “the sum of the A.P. is the cube of its last term”, so

$$(\text{last term})^3=S,$$ i.e.

$${\left(\frac{10}{3}+29d\right)}^{3}=100+435d.$$

Simplifying the equation

Let $$t=\frac{10}{3}+29d$$ (so $$t=l$$). Then $$d=\dfrac{t-\frac{10}{3}}{29}.$$ Substitute this in the right-hand side:

$$t^{3}=100+435\left(\frac{t-\frac{10}{3}}{29}\right).$$

Since $$435/29=15,$$ we get

$$t^{3}=100+15\left(t-\frac{10}{3}\right)=100+15t-50.$$

Thus

$$t^{3}-15t-50=0.$$

Finding the real root

Test small integer values: for $$t=5,$$

$$5^{3}-15\cdot5-50=125-75-50=0.$$

Hence $$t=5$$ is a root. Factorising,

$$(t-5)(t^{2}+5t+10)=0.$$

The quadratic factor has negative discriminant $$25-40=-15,$$ so its roots are complex. Therefore the only real solution is

$$t=5 \quad\Rightarrow\quad l=5.$$

Determining the common difference

Using $$d=\dfrac{t-\frac{10}{3}}{29},$$

$$d=\frac{5-\frac{10}{3}}{29}=\frac{\frac{15}{3}-\frac{10}{3}}{29}=\frac{\frac{5}{3}}{29}=\frac{5}{87}.$$

All 30 terms are non-negative because $$a=\frac{10}{3}\gt0$$ and $$d=\frac{5}{87}\gt0.$$ The progression satisfies every given condition.

Hence the common difference is $$\dfrac{5}{87}$$.

Option A which is: $$\frac{5}{87}$$