The first term of an $AP$ is $-5$ and the last term is $45$. If the sum of the terms of the $AP$ is $120$,then find the number of terms and the common difference.

(A) Let the first term,common difference,and the number of terms of an $AP$ be $a$,$d$,and $n$ respectively.
Given that,first term $(a) = -5$ and last term $(l) = 45$.
Sum of the terms of the $AP = 120 \Rightarrow S_n = 120$.
We know that,if the last term of an $AP$ is known,the sum of $n$ terms is given by:
$S_n = \frac{n}{2}(a + l)$
$120 = \frac{n}{2}(-5 + 45)$
$120 \times 2 = 40 \times n$
$240 = 40n \Rightarrow n = 6$.
Now,to find the common difference,we use the formula for the $n^{th}$ term:
$l = a + (n - 1)d$
$45 = -5 + (6 - 1)d$
$45 + 5 = 5d$
$50 = 5d \Rightarrow d = 10$.
Thus,the number of terms is $6$ and the common difference is $10$.