Question

The first, second and last term of an AP are a, b, c respectievely. Prove that the sum is (a+c)(b+c-2a)/2(b-a)​

Answer 1

Sum = $\dfrac{(a + c) (b + c - 2a)}{b - a}$, it is proved.

Step-by-step explanation:

Here, first term = a,

second term = b and last term = c

⇒ c = a + (n - 1)d     ... (1)

∴ Common difference (d) = Second term  - First term = b - a

We know that,

$S_{n}$ = $\dfrac{n}{2}$·{2a + (n - 1)d   ... (2)

From equation (1), we get

c = a + (n - 1)d  

⇒ c - a =  (n - 1)d

⇒  n - 1 = $\dfrac{c - a}{b - a}$

⇒  n = $\dfrac{c - a}{b - a}$ + 1

∴ n = $\dfrac{ b + c - 2a}{b - a}$

Puttong the value of a and d in equation (2), we get

$S_{n}$ = $\dfrac{(a + c) (b + c - 2a)}{b - a}$

∴ Sum = $\dfrac{(a + c) (b + c - 2a)}{b - a}$, it is proved.