Let the first term be \(a\) and the common ratio be \(r\).

The third term is given by \(ar^2 = 1764\).

The sum of the second and third terms is \(ar + ar^2 = 3444\).

We have the equations:

1. \(ar^2 = 1764\)

2. \(ar + ar^2 = 3444\)

From equation 1, \(ar^2 = 1764\).

From equation 2, \(ar + ar^2 = 3444\), substitute \(ar^2 = 1764\):

\(ar + 1764 = 3444\)

\(ar = 3444 - 1764 = 1680\)

Now, divide the two equations:

\(\frac{ar^2}{ar} = \frac{1764}{1680}\)

\(r = \frac{1764}{1680} = \frac{21}{20} = 1.05\)

Substitute \(r = 1.05\) back into \(ar = 1680\):

\(a \times 1.05 = 1680\)

\(a = \frac{1680}{1.05} = 1600\)

The 50th term is given by \(ar^{49}\):

\(1600 \times (1.05)^{49} \approx 17,500\)