Question

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

Answer 1

It is given that the first term of the arithmetic series is $a_1=a$, the second term is $a_2=b$ and the last term is $T_n=c$

We find the common difference $d$ by subtracting the second term by first term as follows:

$d=b-a$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, with $a=a,d=b-a$ and $T_n=c$, we have

$T_n=a+(n-1)d\Rightarrow c=a+(n-1)(b-a)\Rightarrow c=a+(b-a)n-b+a\Rightarrow c=2a-b+(b-a)n\Rightarrow (b-a)n=c-2a+b\Rightarrow n=\frac{c+b-2a}{b-a}$

We also know that the sum of an arithmetic series with first term $a$ and common difference $d$ is:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now to find the sum of series, substitute

$n=\frac{c+b-2a}{b-a},a=a$ and $d=b-a$

in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

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

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

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

Hence, it is proved that the sum is $\frac{(a+c)(b+c-2a)}{2(b-a)}$.