Question

The first, second and middle terms of an A.P. are $a, b, c,$ respectively. Their sum is

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

Answer 1

Answer: (B)

We have: first term $=a,$ second term $=b \Rightarrow d=$ common difference $=b-a$
It is given that the middle term is $c .$ This means that there are an odd number of terms in the $AP$. Let there be $(2 n+1)$ terms in the $AP$. Then, $(n+1)^{\text {th }}$ term is the middle term.
$\therefore $ middle term $=c$
$\Rightarrow a+n d=c$
$\Rightarrow a+n(b-a)=c$
$\Rightarrow n=\frac{c-a}{b-a}$
$\therefore $ Sum $=\frac{2 n+1}{2}[2 a+(2 n+1-1) d]$
$=(2 n+1)(a+n d)$
$=\left\{2\left(\frac{c-a}{b-a}\right)+1\right\}\left[a+\left(\frac{c-a}{b-a}\right)(b-a)\right]$
$=\frac{2 c(c-a)}{b-a}+c$