If a, b, c, d are in G.P, then 

A)If a, b, c, d are in G.P, then $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$ $(a\ne b\ne c\ne d\ne 0)$

B) a, b, c, d are in G.P

Let $a=a_1\Rightarrow a^n={a_1}^n$

$b=a_{1}r\Rightarrow b^n={a_1}^n.r^n$

$c=a_1{r}^2\Rightarrow c^n={a_1}^n.r^{2n}$

$d=a_1{r}^3\Rightarrow d^n={a_1}^n{r}^{3n}$

$\frac{b^n+c^n}{a^n+b^n}=\frac{{a_1}^n.r^n+{a_1}^n.r^{2n}}{{a_1}^n+{a_1}^n.r^n}=\frac{r^n+r^{2n}}{1+r^n}$

$\frac{c^n+d^n}{b^n+c^n}=\frac{{a_1}^n.r^{2n}+{a_1}^n.r^{3n}}{{a_1}^n.r^n+{a_1}^n.r^{2n}}=\frac{r^n+r^{2n}}{1+r^n}$

Ratio’s are equal

$\therefore a^n+b^n,b^n+c^n,c^n+d^n\text{}areinG.P$

C) $(b-c^2)+{(c-a)}^2+{(d-b)}^2={(a-d)}^2$

$\Rightarrow b^2+c^2-2bc+c^2+a^2-2ac+d^2+b^2-2bd=a^2+d^2-2ab$

$\Rightarrow b^2+c^2-bc-ac-bd=-ad$

$\Rightarrow ac+bd-bc-ac-bd=-ad$

$\Rightarrow -bc=-ad$

$\Rightarrow \frac{b}{a}=\frac{d}{c}$

Therefore above statement is true

D) $\frac{1}{a},\frac{1}{b},\frac{1}{c}and\frac{1}{d}areinG.P$

$\frac{\frac{1}{b}}{\frac{1}{a}}=\frac{\frac{1}{c}}{\frac{1}{b}}=\frac{\frac{1}{d}}{\frac{1}{c}}\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$

$\therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}and\frac{1}{d}areinG.P$