Question

In an A.P. of $n$ terms, $a$ is the first term, $b$ is the second last term and $c$ is the last term, then the sum of all of its term equals

• A. $\frac{(a+c)(a+b-2c)}{2(b-a)}$
• B. $\frac{(a+c)(a+b-c)}{2(c-b)}$
• C. $\frac{(a+c)(a+b-2c)}{(b-c)}$
• D. $\frac{(a+c)(b+c-2a)}{2(a-b)}$

Answer 1

The correct option is A $\frac{(a+c)(a+b-2c)}{2(b-a)}$

We have given that the first term is $a$ and the common difference is $b-a$
Therefore, $a+(n-1)(b-a)=c$
$\Rightarrow (n-1)=\frac{c-a}{b-a}$
$\Rightarrow n=\frac{c-a}{b-a}+1$
$\Rightarrow n=\frac{2c-a-b}{b-a}$
$\Rightarrow n=\frac{a+b-2c}{b-a}$ ...(1)

Now, $S_n=\frac{n}{2}[\text{first term}+\text{last term}]$
${\Rightarrow S}_n=\frac{n}{2}[a+c]$
Substituting the value of $n$ from equation $\left(1\right)$, we get
${\Rightarrow S}_n=\frac{(a+c)(a+b-2c)}{2(b-a)}$