Question

If a, b,c,d are in G.P., then show that
i) $(a+b{)}^2,(b+c{)}^2,(c+d{)}^2$ are in G.P
ii) $\frac{1}{a^2+b^2},\frac{1}{b^2+c^2},\frac{1}{c^2+d^2}$ are in G.P

Answer 1

$a,b,c$ and div GP

Let common ratio=r

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

$\left(i\right)(a+b{)}^2=(a+ar{)}^2=a^2(1+r{)}^{@}(b+c{)}^2=(ar+a{r}^2{)}^2=a^2{r}^2(1+r{)}^2(c+d{)}^2=(a{r}^2+a{r}^3{)}^2=a^2{r}^4(1+r{)}^2$

Here common ratio is $r^2$

and it is in GP Proved

$\left(ii\right)\frac{1}{a^2+b^2}=\frac{1}{a^2+a^2{r}^2}=\frac{1}{a^2(1+r^2)}\frac{1}{b^2+c^2}=\frac{1}{a^2{r}^2+a^2{r}^4}=\frac{1}{a^2{r}^2(1+r^2)}\frac{1}{c^2+d^2}=\frac{1}{a^2{r}^4+a^2{r}^6}=\frac{1}{a^2{r}^4(1+r^2)}$

Here common ratio $=\frac{1}{r^2}$

and it is GP proved