Question

If first term of an A.P is a second term is b and last terms is c, then show that sum of all terms is (a+c)(b+c-2a)
2(b-a) ​

Answer 1

Step-by-step explanation:

Given :

First term of A.P. is = a, Second term = b

Last term = c.

To prove :

$S_n = \frac{(a + c)(b + c - 2a)}{2(b - a)}$

Solution :

Let the last term = nth term of A.P. = c.

Common term d = b - a

Therefore,

$T_n = a + (n - 1)d \\ \implies \: c \: = a \: + (n \: - 1)(b \: - a) \\ \implies \: n \: - 1 \: = \frac{c \: - a}{b \: - a} \\ \implies \: n \: = \frac{c \: - a \: + b \: - a}{b \: - a} \\ \implies \: n \: = \frac{b + c - 2a}{b - a}$

Now, sum of n terms of an A.P. is given by ;

$S_n = \frac{n}{2} [ First \: \: term + last \: \: term] \\ = \frac{(b + c - 2a)}{2(b - a)} (a + c)$

Hence, proved.