The sum of $n$ terms of an $A.P.$ is given by $S_{n} = 5n^{2} - 3n$. Find the $n^{th}$ term of the $A.P.$

(10N-8) Given the sum of $n$ terms: $S_{n} = 5n^{2} - 3n$.
We know that the $n^{th}$ term $T_{n}$ is given by $T_{n} = S_{n} - S_{n-1}$ for $n > 1$.
First,find $S_{n-1}$:
$S_{n-1} = 5(n-1)^{2} - 3(n-1)$
$S_{n-1} = 5(n^{2} - 2n + 1) - 3n + 3$
$S_{n-1} = 5n^{2} - 10n + 5 - 3n + 3$
$S_{n-1} = 5n^{2} - 13n + 8$.
Now,calculate $T_{n}$:
$T_{n} = (5n^{2} - 3n) - (5n^{2} - 13n + 8)$
$T_{n} = 5n^{2} - 3n - 5n^{2} + 13n - 8$
$T_{n} = 10n - 8$.
For $n = 1$,$T_{1} = S_{1} = 5(1)^{2} - 3(1) = 2$.
Using the formula $T_{n} = 10n - 8$,for $n = 1$,$T_{1} = 10(1) - 8 = 2$.
Since both match,the $n^{th}$ term is $T_{n} = 10n - 8$.