Question

Show that the sum of an AP whose first term b and last term c, is equal to
(a+c)(b+c-2a) ÷ 2(b-a)

Answer 1

Step-by-step explanation:

$common \: difference = a _{2} - a _{1} = (b - a) \\ \\ n \: th \: term = a + (n - 1)d \\ c = a + (n - 1)(b - a) \\ n = \frac{b + c - 2a}{ b-a } \\ s_{n } = \frac{n}{2} ( a+ c) = \frac{( b+c -2a )(a +c )}{2(b - a)} \\ hence \: proved$