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Geometric Progressions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions4 Marks
Question
If a, b, c, are in G.P., prove that:
$\big(\text{a}^2-\text{b}^2\big),\big(\text{b}^2-\text{c}^2\big),\big(\text{c}^2-\text{d}^2\big)\text{ are in G.P.}$
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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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