If `a , b ,a n dc` are respectively, the pth, qth , and rth terms of a G.P., show that `(q-r)loga+(r-p)logb+(p-q)logc=0.`

Explanation

Let A be the first term and R the common ratio of the given G.P.
Then, a=pth term
`rArra=AR^(p-1)`
`rArrloga=logA+(p-1)logR`
Similarly,
logb=logA+(q-1)log R
and logc=logA+(r-1)logR
Now, (q-r)loga+(r-p)logb+(p-q)logc
=(q-r){logA+(p-1)logR}+(r-p)
{log A+(q-1)logR}+(p-q){log A+(r-1)logR}
[Using (1),(2) and (3)]
=logA{(q-r)+(r-p)+(p-q)}+log R{(p-1)`xx(q-r)+(q-1)(r-p)+(r-1)(p-q)}`
`=log Axx0+log R{p(q-r)+q(r-p)+r(p-q)-(q-r)-(r-p)-(p-q)}`
=logA`0+logRxx0=0`