Question

If the sum of first $p,q,r$ terms of an A.P are $a,b,c$ respectively. Show that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

Answer 1

Step1: Obtaining the equations for the sum of first $p,q,r$ terms.

The sum of first $n$terms of an A.P. is given by $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$, where $a$ is the first term and $d$ is the common difference.

The sum of first $p$ terms of an A.P. is given by;

$$\begin{gathered} \Rightarrow S_p=\frac{p}{2}\left[2a_1+(p-1)d\right] \\ \Rightarrow a=\frac{p}{2}\left[2a_1+(p-1)d\right](Substituting{S}_{p}asa) \\ \therefore \frac{2a}{p}=2a_1+(p-1)d\left(Multiplyingbothsidesby\frac{2}{p}\right)equation\left(1\right) \end{gathered}$$

The sum of first $q$ terms of an A.P. is given by:

$$\begin{gathered} \Rightarrow S_q=\frac{q}{2}\left[2a_1+(q-1)d\right] \\ \Rightarrow b=\frac{q}{2}\left[2a_1+(q-1)d\right](Substituting{S}_{q}asb) \\ \therefore \frac{2b}{q}=2a_1+(q-1)d\left(Multiplyingbothsidesby\frac{2}{q}\right)equation\left(2\right) \end{gathered}$$

The sum of first $r$ terms of an A.P. is given by:

$$\begin{gathered} \Rightarrow S_r=\frac{r}{2}\left[2a_1+(r-1)d\right] \\ \Rightarrow c=\frac{r}{2}\left[2a_1+(r-1)d\right](Substituting{S}_{r}asc) \\ \therefore \frac{2c}{r}=2a_1+(r-1)d\left(Multiplyingbothsidesby\frac{2}{r}\right)equation\left(3\right) \end{gathered}$$

Step2: Multiply equations (1), (2) and (3) by $\left(q-r\right),\left(r-p\right)$ and $\left(p-q\right)$ respectively and then add.

$$\begin{gathered} \Rightarrow \frac{2a}{p}(q-r)+\frac{2b}{q}(r-p)+\frac{2c}{r}(p-q)=\left(2a_1+(p-1)d\right)(q-r)+\left(2a_1+(q-1)d\right)(r-p)+\left(2a_1+(r-1)d\right)(p-q) \\ \Rightarrow 2\left[\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\right]=2a_1\left[q-r+r-p+p-q\right]+d\left[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)\right]\left(Factorout2fromtheleftside\right) \\ \Rightarrow 2\left[\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\right]=2a_1\left(0\right)+d\left[pq+qr+rp-pq-qr-rp+p-p+q-q+r-r\right] \\ \Rightarrow 2\left[\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\right]=0+d\left(0\right) \\ \Rightarrow 2\left[\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\right]=0+0 \\ \therefore \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\left(Dividebothsidesoftheequationby2\right) \end{gathered}$$

Hence proved.