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Geometric Progressions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions4 Marks
Question
If $\text{x}^{\text{a}}=\text{x}^{\frac{\text{b}}{2}}\text{z}^{\frac{\text{b}}{2}}=\text{z}^{\text{c}},$ then prove that $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P.
✓

Answer

$\text{x}^{\text{a}}=\text{x}^{\frac{\text{b}}{2}}\text{z}^{\frac{\text{b}}{2}}=\text{z}^{\text{c}}=\lambda(\text{ say})$
$\text{x}=\lambda^{\frac{1}{\text{a}}},\text{z}=\lambda^{\frac{1}{\text{c}}}$
$\text{x}^{\frac{\text{b}}{2}}\times\text{z}^{\frac{\text{b}}{2}}=\lambda$
$\lambda^{\frac{1}{\text{a}}\big(\frac{\text{b}}{2}\big)}\times\lambda^{\frac{\text{b}}{2}\times\frac{1}{\text{c}}}=\lambda$
$\lambda^{\frac{\text{b}}{2\text{a}}+\frac{\text{b}}{2\text{c}}}=\lambda^1$
$\frac{\text{b}}{2\text{a}}+\frac{\text{b}}{2\text{c}}=1$
$\frac{1}{\text{a}}+\frac{1}{\text{c}}=\frac{2}{\text{b}}$
$\Rightarrow\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\text{ are in A.P.}$

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