In an $AP$,given $a_{3} = 15$ and $S_{10} = 125$,find $d$ and $a_{10}$.

(N/A) Given that,$a_{3} = 15$ and $S_{10} = 125$.
As the $n^{th}$ term of an $AP$ is $a_{n} = a + (n - 1)d$,
$a_{3} = a + 2d = 15$ $...(i)$
The sum of $n$ terms is $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
$S_{10} = \frac{10}{2}[2a + (10 - 1)d] = 125$
$5(2a + 9d) = 125$
$2a + 9d = 25$ $...(ii)$
Multiplying equation $(i)$ by $2$,we get:
$2a + 4d = 30$ $...(iii)$
Subtracting equation $(iii)$ from $(ii)$:
$(2a + 9d) - (2a + 4d) = 25 - 30$
$5d = -5$
$d = -1$
Substituting $d = -1$ in equation $(i)$:
$a + 2(-1) = 15$
$a - 2 = 15$
$a = 17$
Now,$a_{10} = a + (10 - 1)d$
$a_{10} = 17 + 9(-1)$
$a_{10} = 17 - 9 = 8$.
Thus,$d = -1$ and $a_{10} = 8$.