If p,q and r (`pneq`) are terms ( not necessarily consecutive) of an A.P., then prove that there exists a rational number k such that `(r-q)/(q-p)`=k. hence, prove that the numbers `sqrt2,sqrt3 and sqrt5` cannot be the terms of a single A.P. with non-zero common difference.

Explanation

Let p,q,r be the lth,mth and nth terms, respectively of the A.P., Then
p=a+(l-1)d,
q=a+(m-1)d
and r=a+(n-1)d.
`therefore` r-q=(n-m)d
and q-p=(m-l)d
`rArr (r-q)/(p-q)=((n-m)d)/((m-l)d)=(n-m)/(m-l)(becausedne0)` …(1)
Since l,m ,n are +ve integers and `mnel,(n-m)/(m-l)` is a rational number.
Substituting `p=sqrt2,q=sqrt3andr=sqrt5` into (1), we get
`(sqrt5-sqrt3)/(sqrt2-sqrt3)==(n-m)/(m-l)`
This is not possible as LHS is rational.
Hence `sqrt2,sqrt3 andsqrt5` cannot be the terms of an A.P.