If the sum of p, q, r terms of an AP are a, b, c respectively, show that If the sum of p, q, r terms of an AP are a, b, c respectively, show that 
$(q-r)$$\frac ap$+$(r-p)$$\frac bq$+$(p-q)$$\frac cr$$=$$0$
 A: i) Let the first term be t, common difference be d 
ii) So sum to p terms is: Sp = (p/2){2t + (p - 1)d} = a 
So, 2a/p = 2t + pd - d 
==> 2(a/p)*(q - r) = (2t + pd - d)(q - r) = 2qt - 2rt + pqd - prd - qd + rd ----- (1) 
iii) Similarly, 2*(b/q)(r - p) = 2rt - 2pt + qrd - pqd - rd + pd ----- (2) 
and 2(c/r)*(p - q) = 2pt - 2qt + prd - qrd - pd + qd ---- (3) 
iv) Adding (1), (2) & (3): 
2*{(a/p)(q - r) + (b/q)(r - p) + (c/r)*(p - q)} = 0 
==> (a/p)(q - r) + (b/q)(r - p) + (c/r)*(p - q) = 0 
A: 
Q:The sum of the first $p, q, r$ terms of an A.P. are $a, b, c$ respectively. Show that
$$
\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0
$$

Sol. Let $x$ be the first term and $d$ the common difference. Then,
$$
\begin{aligned}
S_{p}=a & \Rightarrow\left(\frac{p}{2}\right) \cdot[2 x+(p-1) d]=a \\
& \Rightarrow \frac{a}{p}=x+(p-1) \cdot \frac{d}{2} \\
& \Rightarrow \frac{a(q-r)}{p}=x(q-r)+(p-1)(q-r) \cdot \frac{d}{2}
\end{aligned}
$$
Similarly,
$$
S_{q}=b \Rightarrow \frac{b(r-p)}{q}=x(r-p)+(q-1)(r-p)-\frac{d}{2}
$$
(ii)
And $S_{r}=c \Rightarrow \frac{c(p-q)}{r}=x(p-q)+(r-1)(p-q) \cdot \frac{d}{2}$
(iii
Adding (i), (ii) and (iii), we get $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
