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Geometric Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSGeometric Progressions3 Marks
Question
If $\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}=\frac{\text{b}+\text{cx}}{\text{b}-\text{cx}}=\frac{\text{c}+\text{dx}}{\text{c}-\text{dx}}(\text{x}\neq0),$ then show that a, b, c and d are in G.P.
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Answer

$\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}=\frac{\text{b}+\text{cx}}{\text{b}-\text{cx}}=\frac{\text{c}+\text{dx}}{\text{c}-\text{dx}}$ Now, $\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}=\frac{\text{b}+\text{cx}}{\text{b}-\text{cx}}$ Applying componendo and dividendo $\Rightarrow\frac{(\text{a}+\text{bx})+(\text{a}-\text{bx})}{(\text{a}+\text{bx})-(\text{a}-\text{bx})}=\frac{(\text{b}+\text{cx})+(\text{b}-\text{cx})}{(\text{b}+\text{cx})-(\text{b}-\text{cx})}$ $\Rightarrow\frac{2\text{a}}{2\text{bx}}=\frac{2\text{b}}{2\text{cx}}$ $\Rightarrow\frac{\text{a}}{\text{b}}=\frac{\text{b}}{\text{c}}$ Similiarly, $\frac{(\text{b}+\text{cx})+(\text{b}-\text{cx})}{(\text{b}+\text{cx})-(\text{b}-\text{cx})}=\frac{(\text{c}+\text{dx})+(\text{c}-\text{dx})}{(\text{c}+\text{dx})-(\text{b}-\text{dx})}$ $\Rightarrow\frac{\text{b}}{\text{c}}=\frac{\text{c}}{\text{d}}$ $\therefore$ a, b, c and d are in G.P.

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