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

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 $(2n+1)$ terms in the $AP$. Then, $(n+1{)}^{\text{th }}$ term is the middle term.
$\therefore$ middle term $=c$
$\Rightarrow a+nd=c$
$\Rightarrow a+n(b-a)=c$
$\Rightarrow n=\frac{c-a}{b-a}$
$\therefore$ Sum $=\frac{2n+1}{2}[2a+(2n+1-1)d]$
$=(2n+1)(a+nd)$
$=\left\{2\left(\frac{c-a}{b-a}\right)+1\right\}\left[a+\left(\frac{c-a}{b-a}\right)(b-a)\right]$
$=\frac{2c(c-a)}{b-a}+c$