Question

Show that the sum of an AP whose first term is $a$, the second term $b$ and the last term is $c$ is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}$

Answer 1

First term $=a$

second term $=b$ Last term $=c$

Difference $d=b-a$

In an $AP,n^{th}$ term $T_n=a+(n-1)d$

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

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

$\therefore c-a=(n-1)(b-a)$

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

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

Sum of $n$ numbers $=n/2$ ( first term+last term)

$=n/2(a+c)$

$=\frac{(\frac{c-a}{b-a}+1)(a+c)}{2}$

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

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