The \(p\)th term of a progression is \(P\), the \(q\)th term is \(Q\), and the \(r\)th term is \(R\). Show that, if the progression is arithmetical, \[P(q-r)+Q(r-p)+R(p-q)=0\]

Let \(a\) be the first term of the progression and \(d\) be the common difference. Then \[P=a+(p-1)d, \quad Q=a+(q-1)d\quad\text{and }R=a+(r-1)d.\]

Therefore, \[\begin{align*} P(q-r)+&Q(r-p)+R(p-q)\\ &=\bigl(a+(p-1)d\bigr)(q-r)+\bigl(a+(q-1)d\bigr)(r-p)+\bigl(a+(r-1)d\bigr)(p-q)\\ &=a(q-r+r-p+p-q)+d\bigl((p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)\bigr)\\ &=d(pq-pr-q+r+qr-qp-r+p+rp-qr-p+q)\\ &=0. \end{align*}\]Alternatively, \(P-Q = (p-q)d\), and \(P-R = (p-r)d\). We can divide these to eliminate \(d\), and then rearrange to get the same result.

…and that, if it is geometrical, \[(q-r)\log P+(r-p)\log Q+(p-q)\log R=0.\]

Let \(a\) be the first term of the progression and \(r\) be the common ratio.

Then, \[P=ar^{p-1},\quad Q=ar^{q-1}\quad\text{and}\quad R=ar^{r-1}.\]

Therefore, taking logs,

\[\begin{align*} \log P&=\log a+(p-1)\log r,\\ \log Q&=\log a+(q-1)\log r,\\ \log R&=\log a+(r-1)\log r. \end{align*}\]We can now see that \(\log P, \log Q\) and \(\log R\) are the \(p^{th}\), \(q^{th}\) and \(r^{th}\) terms in an arithmetic progression with common difference \(\log r\).

So using the first part of the question, \((q-r)\log P+(r-p)\log Q+(p-q)\log R = 0\).