Question

Prove that if p, q, r $(p\ne q)$ are terms (not necessarily consecutive) of an A.P., then there exists a rational number k such that $(r-q)/(q-p)=k$.

Answer 1

Let p,q,r be the lth, mth and nth terms of an A.P., then

$p=a+(l-1)d,q=a+(m+1)d$

and $r=a+(n-1)d$

when $r-q=(n-m)d$

and $q-p=m(m-l)d$

so that $\frac{r-q}{q-p}=\frac{(n-m)d}{(m-l)d}=\frac{n-m}{m-l}$ $(\because d\ne 0)$

Since l, m, n are $+$ive integers and $m\ne l$, $(n-m)/(m-1l)$ is a rational number.