# If a,b and c are the p^(th), q^(th) and r^(th) term of an AP then show that p(b-c)+q(c-a)+q(a-b)=0?

May 31, 2018

See below

#### Explanation:

Note there is a typo in the question:

$p \left(b - c\right) + q \left(c - a\right) + \textcolor{red}{r} \left(a - b\right) = 0$

The AP, with common difference $\delta$ and first term $\alpha$, is:

• $\alpha + \left(\alpha + \delta\right) + \ldots . . + {\underbrace{\left(\alpha + \left(p - 1\right) \delta\right)}}_{a = \text{p-th term")+ ..... + underbrace((alpha + (q-1) delta ))\_(b = "q-th term")+ ....... + underbrace((alpha + (r-1) delta))\_(c = "r-th term}} + \ldots \ldots$

$\boldsymbol{\implies} p \left(b - c\right) + q \left(c - a\right) + r \left(a - b\right)$

$= p \left(\alpha + \left(q - 1\right) \delta - \alpha - \left(r - 1\right) \delta\right) + q \left(\alpha + \left(r - 1\right) \delta - \alpha - \left(p - 1\right) \delta\right) + r \left(\alpha + \left(p - 1\right) \delta - \alpha - \left(q - 1\right) \delta\right)$

Eliminate $\alpha$'s:

$= p \left(\left(q - 1\right) \delta - \left(r - 1\right) \delta\right) + q \left(\left(r - 1\right) \delta - \left(p - 1\right) \delta\right) + r \left(\left(p - 1\right) \delta - \left(q - 1\right) \delta\right)$

Factor out $\delta$:

$= \delta \left(p \left(\left(q - 1\right) - \left(r - 1\right)\right) + q \left(\left(r - 1\right) - \left(p - 1\right)\right) + r \left(\left(p - 1\right) - \left(q - 1\right)\right)\right)$

Expand:

$= \delta \left(p \left(q - r\right) + q \left(r - p\right) + r \left(p - q\right)\right)$

$= \delta \left(\textcolor{red}{p q} - \textcolor{b l u e}{p r} + \boldsymbol{q r} - \textcolor{red}{q p} + \textcolor{b l u e}{r p} - \boldsymbol{r q}\right) \boldsymbol{\textcolor{g r e e n}{= 0}}$