Question

If first term of an A.P. is $a$, second term is $b$ and the last term is $c$, than show that the sum of all term is $\frac{\left(a+c\right)\left(b+c-2a\right)}{2\left(b-a\right)}$.

Answer 1

Given first term of AP$=$a

Given second term of AP$=$b

Given third term of AP$=$c

Common difference of AP$=b-a$

Now, $a_n=$last term$=a+(n-1)d$

$c=a+(n-1)(b-a)$

$c-a+b-a=n$

______________

$b-a$

______________

$\frac{b+c-2a}{b-a}=n$

Now, sum of AP$=S_n=\frac{n}{2}[a+a_n]$

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