Question

If $a,b,c,d$ are in $G.P$,prove that
$\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

Answer 1

It is given that $a,b,c,d$ are in G.P., then,

$a=a$, $b=ar$, $c=a{r}^2$ and $d=a{r}^3$.

Putting the values on the LHS of $\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

$\frac{ab-cd}{b^2-c^2}=\frac{a\left(ar\right)-\left(a{r}^2\right)\left(a{r}^3\right)}{{\left(ar\right)}^2-{\left(a{r}^2\right)}^2}$

$=\frac{a^{2}r-a^2{r}^5}{a^2{r}^2-a^2{r}^4}$

$=\frac{a^{2}r\left(1-r^4\right)}{a^2{r}^2\left(1-r^2\right)}$

$=\frac{{\left(1\right)}^2-{\left(r^2\right)}^2}{r\left(1-r^2\right)}$

$=\frac{\left(1-r^2\right)\left(1+r^2\right)}{r\left(1-r^2\right)}$

$=\frac{1+r^2}{r}$

Now putting the values in RHS of $\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

$\frac{a+c}{b}=\frac{a+a{r}^2}{ar}$

$=\frac{a\left(1+r^2\right)}{ar}$

$=\frac{1+r^2}{r}$

Therefore, $LHS=RHS$.

Hence proved.