Question

If $a$, $b$ and $c$ are the $1\mathrm{st},3\mathrm{rd}$ $n\mathrm{th}$ terms respectively of an AP, then prove that the sum to $n$ terms is $\frac{c+a}{2}+\frac{c^2-a^2}{b-a}$.

Answer 1

Step 1: Find the expression for $\mathit{n}$:

Formula:

The $n^{\mathrm{th}}$ term of an AP is $a_n=a+\left(n-1\right)d$.

Let the first term of the given AP be $a$ and its common difference be $d$.

According to the question, we have $a_1=a,a_3=b$ and $a_n=c$. We know that $a_3=a+2d$, so we can write $a+2d=b$.

Adding $a_1$ and $a_3$ we get,

$$\begin{gathered} \Rightarrow a_1+a_3=a+\left(a+2d\right) \\ \Rightarrow a+b=2a+2d \\ \Rightarrow b-a=2d \\ \Rightarrow d=\frac{b-a}{2} \end{gathered}$$

Substitute $d=\frac{b-a}{2}$ into the $n^{\mathrm{th}}$ term of an AP formula.

$$\begin{gathered} \Rightarrow c=a+(n-1)\frac{b-a}{2}\left[\because a_n=c\right] \\ \Rightarrow c-a=\left(n-1\right)\frac{b-a}{2} \\ \Rightarrow \frac{2\left(c-a\right)}{b-a}=n-1 \\ \Rightarrow \frac{2\left(c-a\right)+b-a}{b-a}=n \end{gathered}$$

Step 2: Compute the sum of $\mathit{n}$ terms:

Formula:

The sum of $n$ terms of a series in an AP is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.

Substitute $d=\frac{b-a}{2}$, $n=\frac{2\left(c-a\right)}{b-a}$ and $n=\frac{2(c-a)+b-a}{b-a}$ into the formula for $S_n$.

$\begin{array}{rcl}& \Rightarrow & S_n=\frac{2\left(c-a\right)+b-a}{2(b-a)}\left[2a+\frac{2\left(c-a\right)}{\left(b-a\right)}\left(\frac{\left(b-a\right)}{2}\right)\right]\\ & =& \frac{2\left(c-a\right)+b-a}{2\left(b-a\right)}\left[\left(a+c\right)\right]\\ & =& \frac{2\left(c-a\right)\left(c+a\right)}{2\left(b-a\right)}+\frac{\left(b-a\right)\left(c+a\right)}{2\left(b-a\right)}\\ & =& \frac{c+a}{2}+\frac{c^2-a^2}{b-a}\end{array}$

Hence, proved.