Show that the sum of an $AP$ whose first term is $a,$ the second term is $b,$ and the last term is $c,$ is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}.$

(A) Given that the $AP$ is $a, b, \dots, c.$
Here,the first term $= a,$ and the common difference $d = b - a.$
The last term is $l = a_n = c.$
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d.$
Substituting the values: $c = a + (n - 1)(b - a).$
$(n - 1) = \frac{c - a}{b - a}.$
$n = \frac{c - a}{b - a} + 1 = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \dots (i).$
The sum of an $AP$ is given by $S_n = \frac{n}{2}(a + l).$
Substituting $n$ from $(i)$ and $l = c:$
$S_n = \frac{1}{2} \left( \frac{b + c - 2a}{b - a} \right) (a + c).$
Thus,$S_n = \frac{(a + c)(b + c - 2a)}{2(b - a)}.$ Hence proved.