Certainly! Here's the solution with \(\) instead of \[\] for the MathJax:

To find the sum of the first four terms of a geometric progression, we first need to determine the common ratio \(r\) of the progression.

We are given:
First term (\(a\)) = 153
Sixth term = \(\frac{17}{27}\)

The nth term of a geometric progression is given by:
\(a_n = a \cdot r^{(n-1)}\)

Using the information provided, we can set up two equations:

For the first term: 
\(153 = a \cdot r^{(1-1)} = a\)

For the sixth term:
\(\frac{17}{27} = 153 \cdot r^{(6-1)}\)

Let's solve the second equation for \(r\):

\(\frac{17}{27} = 153 \cdot r^5\)

Now, isolate \(r\):

\(r^5 = \frac{17}{27 \cdot 153}\)

Taking the fifth root of both sides:

\(r = \left(\frac{17}{27 \cdot 153}\right)^{\frac{1}{5}}\)

Now that we have found \(r\), we can find the sum of the first four terms using the formula for the sum of a geometric progression:

\(S_n = \frac{a(1 - r^n)}{1 - r}\)

In this case, \(n\) is 4 (the sum of the first four terms), \(a\) is 153, and we have found the value of \(r\).

Now, plug in the values:

\(S_4 = \frac{153(1 - \left(\frac{17}{27 \cdot 153}\right)^{\frac{4}{5}})}{1 - \left(\frac{17}{27 \cdot 153}\right)^{\frac{1}{5}}}\)

Now, calculate this expression to find the sum:

\(S_4 \approx \frac{680}{3}\)

So, the sum of the first four terms is approximately \(\frac{680}{3}\).