In an $AP$,given $d=5$ and $S_{9}=75$,find $a$ and $a_{9}$.

(N/A) Given that,$d = 5$ and $S_{9} = 75$.
The formula for the sum of $n$ terms of an $AP$ is $S_{n} = \frac{n}{2}[2a + (n-1)d]$.
Substituting the given values for $n=9$:
$75 = \frac{9}{2}[2a + (9-1)5]$
$75 = \frac{9}{2}[2a + 40]$
$75 = 9(a + 20)$
Dividing both sides by $3$:
$25 = 3(a + 20)$
$25 = 3a + 60$
$3a = 25 - 60$
$3a = -35$
$a = -\frac{35}{3}$
Now,to find the $9^{th}$ term $(a_{9})$,we use the formula $a_{n} = a + (n-1)d$:
$a_{9} = a + (9-1)d$
$a_{9} = -\frac{35}{3} + 8(5)$
$a_{9} = -\frac{35}{3} + 40$
$a_{9} = \frac{-35 + 120}{3}$
$a_{9} = \frac{85}{3}$