Question

If the ${\mathrm{p}}^{\mathrm{th}}$, ${\mathrm{q}}^{\mathrm{th}}$ and ${\mathrm{r}}^{\mathrm{th}}$ terms of a GP are $a,b$and $c$ respectively. Prove that ${\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}=1$

Answer 1

To prove ${\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}=1$:

LHS$={\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}$

Let $A$be the first term and $R$be the common term, then we have

$\begin{array}{cc}\Rightarrow & a=A{R}^{\left(p-1\right)}\\ \Rightarrow & b=A{R}^{\left(q-1\right)}\\ \Rightarrow & c=A{R}^{\left(r-1\right)}\end{array}$

Now, substitute the value of $a,b$ and $c$ in ${\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}$

$\begin{array}{c}\begin{array}{c}\Rightarrow {\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}={\left[{\mathrm{AR}}^{(\mathrm{p}-1)}\right]}^{(\mathrm{q}-\mathrm{r})}{\left[{\mathrm{AR}}^{(\mathrm{q}-1)}\right]}^{(\mathrm{r}-\mathrm{p})}{\left[{\mathrm{AR}}^{(\mathrm{r}-1)}\right]}^{(\mathrm{p}-\mathrm{q})}\\ \Rightarrow {\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}={\mathrm{A}}^{(\mathrm{q}-\mathrm{r})}{\mathrm{R}}^{(\mathrm{p}-1)(\mathrm{q}-\mathrm{r})}{\mathrm{A}}^{(\mathrm{r}-\mathrm{p})}{\mathrm{R}}^{(\mathrm{q}-1)(\mathrm{r}-\mathrm{p})}{\mathrm{A}}^{(\mathrm{p}-\mathrm{q})}{\mathrm{R}}^{(\mathrm{r}-1)(\mathrm{p}-\mathrm{q})}\\ \Rightarrow {\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}={\mathrm{A}}^0{\mathrm{R}}^0\\ \Rightarrow {\mathrm{a}}^{\mathrm{q}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}-\mathrm{p}}{\mathrm{c}}^{\mathrm{p}-\mathrm{q}}=1\end{array}\end{array}$

$\therefore$LHS=RHS

Hence proved.