The problem provides an arithmetic progression (AP) given by $k, \frac{2k}{3}, \frac{k}{3}, 0, ...$. We are asked to find (i) the sixth term of the AP, and (ii) a general expression for the $n^{th}$ term of the AP.

1. Problem Description

The problem provides an arithmetic progression (AP) given by $k, \frac{2k}{3}, \frac{k}{3}, 0, ...$. We are asked to find (i) the sixth term of the AP, and (ii) a general expression for the $n^{th}$ term of the AP.

2. Solution Steps

First, we need to find the common difference, $d$, of the AP.

$d = a_2 - a_1 = \frac{2k}{3} - k = \frac{2k - 3k}{3} = -\frac{k}{3}$

$d = a_3 - a_2 = \frac{k}{3} - \frac{2k}{3} = -\frac{k}{3}$

The common difference is $d = -\frac{k}{3}$.

(i) The general formula for the $n^{th}$ term of an AP is:

$a_n = a_1 + (n-1)d$

Here, $a_1 = k$ and $d = -\frac{k}{3}$.

We want to find the sixth term, so $n=6$.

$a_6 = a_1 + (6-1)d = k + 5(-\frac{k}{3}) = k - \frac{5k}{3} = \frac{3k - 5k}{3} = -\frac{2k}{3}$

(ii) Now, we need to find the $n^{th}$ term.

$a_n = a_1 + (n-1)d = k + (n-1)(-\frac{k}{3}) = k - \frac{k(n-1)}{3} = \frac{3k - kn + k}{3} = \frac{4k - kn}{3} = \frac{k(4-n)}{3}$

3. Final Answer

(i) The sixth term is $-\frac{2k}{3}$.

(ii) The $n^{th}$ term is $\frac{k(4-n)}{3}$.