The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 5 - 6n$. Find the sum of the first $n$ terms of the $A.P.$

(N/A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 5 - 6n$.
To find the first term $(a)$,substitute $n = 1$:
$a = T_{1} = 5 - 6(1) = -1$.
To find the second term $(T_{2})$,substitute $n = 2$:
$T_{2} = 5 - 6(2) = 5 - 12 = -7$.
The common difference $(d)$ is $T_{2} - T_{1} = -7 - (-1) = -6$.
The sum of the first $n$ terms $(S_{n})$ is given by the formula $S_{n} = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values of $a = -1$ and $d = -6$:
$S_{n} = \frac{n}{2} [2(-1) + (n - 1)(-6)]$
$S_{n} = \frac{n}{2} [-2 - 6n + 6]$
$S_{n} = \frac{n}{2} [4 - 6n]$
$S_{n} = n(2 - 3n) = -3n^{2} + 2n$.