Review question

# How are the $p$th, $q$th and $r$th terms of an AP related? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6137

## Solution

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, $(q-r)\log P+(r-p)\log Q+(p-q)\log R = 0$.