If $$\left(2 + \frac{x}{3}\right)^{55}$$ is expanded in the ascending powers of x and the coefficients of powers of x in two consecutive terms of the expansion are equal, then these terms are:

In the binomial expansion of $$(a + b)^n$$, the general term $$T_{r+1}$$ is given by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

The $$(r+1)^{th}$$ term is $$T_{r+1} = \binom{55}{r} (2)^{55-r} \left(\frac{x}{3}\right)^r = \left[ \binom{55}{r} 2^{55-r} \left(\frac{1}{3}\right)^r \right] x^r$$

The coefficient of the power $$x^r$$ is $$C_r = \binom{55}{r} 2^{55-r} 3^{-r}$$

We are looking for two consecutive terms with equal coefficients. Let these be the coefficients of $$x^r$$ and $$x^{r+1}$$. We set $$C_r = C_{r+1}$$:

$$\binom{55}{r} 2^{55-r} 3^{-r} = \binom{55}{r+1} 2^{55-(r+1)} 3^{-(r+1)}$$

$$\binom{55}{r} \cdot 2 = \binom{55}{r+1} \cdot \frac{1}{3}$$

$$6 \cdot \binom{55}{r} = \binom{55}{r+1}$$

Using the property $$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}$$:

$$6 = \frac{55 - r}{r + 1}$$

$$6(r + 1) = 55 - r$$

$$6r + 6 = 55 - r$$

$$7r = 49 \implies \mathbf{r = 7}$$

Since $$r = 7$$, the powers of $$x$$ with equal coefficients are $$x^7$$ and $$x^8$$:

The term containing $$x^7$$ is the $$(7+1)^{th} = \mathbf{8^{th}}$$ term.

The term containing $$x^8$$ is the $$(8+1)^{th} = \mathbf{9^{th}}$$ term.