Question

Show that the sum of an A.P.whose first term is a, the second term is b and the
last term is c, is equal to (a+c)(b+c-2a)/2(b-a).​

Answer 1

Step-by-step explanation:

Given:

$t_1 = a, \: t_2 = b, \: t_n = c \\ \therefore \: d = b - a \\ \because \: t_n = a + (n - 1)d \\ \therefore \: c = a + (n - 1)(b - a) \\ \therefore \: c - a = (n - 1)(b - a) \\ \frac{c - a}{b - a} = n - 1 \\ \implies \: n = \frac{c - a}{b - a} + 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{c - a + b - a}{b - a} \\ \implies \fbox {\: n = \frac{b + c - 2a}{b - a}} \\ now \: sum \: of \: AP \\S_n = \frac{n}{2}(a+t_n)\\</p><p>= \frac{\frac{(b + c - 2a)}{(b - a)}}{2}(a+c)\\</p><p>\fbox{\therefore S_n= \frac{(a+c)(b + c - 2a)}{2(b - a)}}$

Thus proved.