Show that the sum of an AP

Solution:

Given that, the AP is a, b………..c

Here, first term = a, common difference = b – a

and last term, $l=a_{n}=c$

$\because$ $a_{n}=l=a+(n-1) d$

$\Rightarrow \quad c=a+(n-1)(b-a)$

$\Rightarrow \quad(n-1)=\frac{c-a}{b-a}$

$\Rightarrow \quad n=\frac{c-a}{b-a}+1$

$\Rightarrow \quad n=\frac{c-a+b-a}{b-a}=\frac{c+b-2 a}{b-a}$ $\ldots$ (i)

$\therefore$ Sum of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{(b+c-2 a)}{2(b-a)}\left[2 a+\left\{\frac{b+c-2 a}{b-a}-1\right\}(b-a)\right]$

 

$=\frac{(b+c-2 a)}{2(b-a)}\left[2 a+\frac{c-a}{b-a} \cdot(b-a)\right]$

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

 

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

Hence proved.