Write down the value of \(T_4\).

\[T_1 = -(t^2 + 6t + 9) = -(t + 3)^2\]\[T_2 = \frac{(t + 3)^3}{2}\]\[r = \frac{T_2}{T_1} = \frac{\frac{(t + 3)^3}{2}}{-(t + 3)^2} = -\frac{t + 3}{2}\]

For convergence:

\[-1 \lt r \lt 1\]\[-1 \lt -\frac{t + 3}{2} \lt 1\]\[-2 \lt -t - 3 \lt 2\]\[1 \lt -t \lt 5\]

\(-5 \lt t \lt -1\) (with \(t \neq -3\), since the terms cannot be zero)

First differences: \(28\,;\,20\,;\,12\,;\,\ldots\) is arithmetic with \(a = 28\) and \(d = -8\)

General term of first differences:

\[28 + (n - 1)(-8) = 36 - 8n\]

12th term:

\[36 - 8(12) = -60\]\[-4n^2 + 40n + 36 = -60\]\[-4n^2 + 40n + 96 = 0\]\[n^2 - 10n - 24 = 0\]\[(n - 12)(n + 2) = 0\]

\(n = 12\) (\(n = -2\) is not valid)

It is the 12th term (\(n = 12\)).

\[S_{\infty} = \frac{a}{1 - r} = \frac{8}{1 - \left(-\frac{1}{2}\right)}\]\[= \frac{8}{\frac{3}{2}}\]

\(S_{\infty} = \frac{16}{3} \approx 5{,}33\)

\[S_{\infty} = \frac{a}{1 - r} = 27\]\[S_3 = \frac{a(1 - r^3)}{1 - r} = 26\]

Substitute \(\frac{a}{1 - r} = 27\):

\[27(1 - r^3) = 26\]\[1 - r^3 = \frac{26}{27}\]\[r^3 = \frac{1}{27}\]

\(r = \frac{1}{3}\)

Number of terms:

\[n = 117 - k + 1 = 118 - k\]

First term: \(4k - 1\); last term: \(4(117) - 1 = 467\)

\[S_n = \frac{n}{2}(a + l)\]\[26\,675 = \frac{118 - k}{2}(4k - 1 + 467)\]\[26\,675 = (118 - k)(2k + 233)\]\[26\,675 = 236k + 27\,494 - 2k^2 - 233k\]\[2k^2 - 3k - 819 = 0\]\[(k - 21)(2k + 39) = 0\]

\(k = 21\) or \(k = -\frac{39}{2}\) (not a whole number)

\(k = 21\)

\[T_{50} = 6(50)^2 - 9(50)\]\[= 15\,000 - 450\]

\(T_{50} = 14\,550\)