If a, b, c, d are in GP, prove that

To prove: $\left(a^{2}-b^{2}\right),\left(b^{2}-c^{2}\right),\left(c^{2}-d^{2}\right)$ are in GP.

Given: $a, b, c$ are in GP

Formula used: When $a, b, c$ are in $G P, b^{2}=a c$

Proof: When $a, b, c, d$ are in GP then

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

From the above, we can have the following conclusion

$\Rightarrow \mathrm{bc}=\mathrm{ad} \ldots$ (i)

$\Rightarrow \mathrm{b}^{2}=\mathrm{ac} \ldots$ (ii)

$\Rightarrow \mathrm{c}^{2}=\mathrm{bd} \ldots$ (iii)

Considering $\left(a^{2}-b^{2}\right),\left(b^{2}-c^{2}\right),\left(c^{2}-d^{2}\right)$

$\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=a^{2} c^{2}-a^{2} d^{2}-b^{2} c^{2}+b^{2} d^{2}$

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

From eqn. (i) , (ii) and (iii)

$=\left(b^{2}\right)^{2}-(b c)^{2}-(b c)^{2}+\left(c^{2}\right)^{2}$

$=b^{4}-2 b^{2} c^{2}+c^{4}$

$\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=\left(b^{2}-c^{2}\right)^{2}$

From the above equation we can say that $\left(a^{2}-b^{2}\right),\left(b^{2}-c^{2}\right),\left(c^{2}-d^{2}\right)$ are in GP