Question
If a, b, c, are in G.P., prove that: $\big(\text{a}^2-\text{b}^2\big),\big(\text{b}^2-\text{c}^2\big),\big(\text{c}^2-\text{d}^2\big)\text{ are in G.P.}$

Answer

a, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ad}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ Now, $\big(\text{b}^2-\text{c}^2\big)^2=\big(\text{b}^2\big)-2\text{b}^2\text{c}^2+\big(\text{c}^2\big)^2$ $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\big(\text{ac}\big)^2-\text{b}^2\text{c}^2-\text{b}^2\text{c}^2+\big(\text{bd}\big)^2$ [Using (1)] $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\text{a}^2\text{c}^2-\text{b}^2\text{c}^2-\text{a}^2\text{d}^2+\text{b}^2\text{d}^2$ [Using (1)] $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\text{c}^2(\text{a}^2-\text{b}^2\big)-\text{d}^2\big(\text{a}^2-\text{b}^2\big)$ $\Rightarrow\big(\text{b}^2-\text{c}^2\big)^2=\big(\text{a}^2-\text{b}^2\big)\big(\text{c}^2-\text{d}^2\big)$ $\therefore\big(\text{a}^2-\text{b}^2\big),\big(\text{b}^2-\text{c}^2\big)\text{ and }\big(\text{c}^2-\text{d}^2\big)\text{ are also in G.P.}$

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