Question
If a, b, c are in G.P., prove that:
$\frac{1}{\text{a}^2-\text{b}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{b}^2-\text{c}^2}$
$\frac{1}{\text{a}^2-\text{b}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{b}^2-\text{c}^2}$
a, b = ar, c = ar2
$\text{L.H.S}=\frac{1}{\text{a}^2-\text{b}^2}+\frac{1}{\text{b}^2}$
$=\frac{1}{\text{a}^2-\text{a}^2\text{r}^2}+\frac{1}{\text{a}^2\text{r}^2}$
$=\frac{1}{\text{a}^2}\Big[\frac{1}{1-\text{r}^2}+\frac{1}{\text{r}^2}\Big]$
$=\frac{1}{\text{a}^2}\Bigg[\frac{\text{r}^2+1-\text{r}^2}{\big(1-\text{r}^2\big)\text{r}^2}\Bigg]$
$=\frac{1}{\text{a}^2}\Big[\frac{1}{\text{r}^2-\text{r}^4}\Big]$
$=\frac{1}{(\text{ar})^2-(\text{ar}^2)^2}$
$=\frac{1}{\text{b}^2-\text{c}^2}$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
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