Question

If $a,b,c$ are the $pth,qth$ and $rth$ terms of a $GP$, show that $(q-r)\log a+(r-p)\log b+(p-q)\log c=0$.

Answer 1

Let A be the first term and R be the common ratio of G.P

A/Q, $a=A{R}^{p-1}b=A{R}^{q-1}c=A{R}^{r-1}$

now, $\log a=\log A{R}^{p-1}\Rightarrow \log a=\log A+(p-1)\log R$

similarly, $\log b=\log A+(q-1)\log R\log c=\log A+(r-1)\log R$

Putting the above value in the equation, we get;

$LHS=(q-r)\log a+(r-p)\log b+(p-q)\log c=(q-r)[\log A+(p-1\left)\log R\right]+(r-p)[\log A+(q-1\left)\log R\right]+(p-q)[\log A+(r-1\left)\log R\right]=\log A\left[\right(q-r)+(r-p)+(p-q\left)\right]+\log R\left[\right(q-r\left)\right(p-1)+(r-p\left)\right(q-1)+(p-q\left)\right(r-1\left)\right]=0.\log A+0.\log R=0=RHS$