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Geometric Progressions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions3 Marks
Question
If $a, b, c$ are in G.P., prove that:
$\text{a}^2\text{b}^2\text{c}^2\Big(\frac{1}{\text{a}^3}+\frac{1}{\text{b}^3}+\frac{1}{\text{c}^2}\Big)=\text{a}^3+\text{b}^3+\text{c}^3$
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Answer

$a, b, c$ are in G.P.
$a, b = ar, c = ar^2$​​​​​​​
$\text{L.H.S}=\text{a}^2\text{b}^2\text{c}^2\Big(\frac{1}{\text{a}^3}+\frac{1}{\text{b}^3}+\frac{1}{\text{c}^3}\Big)$
$=\text{a}^2\times\text{a}^2\text{r}^2\times\text{a}^2\text{r}^4\Big(\frac{1}{\text{a}^3}+\frac{1}{\text{a}^3\text{r}^3}+\frac{1}{\text{a}^3\text{r}^6}\Big)$
$=\text{a}^6\text{r}^6\Big(\frac{\text{r}^6+\text{r}^3+1}{\text{a}^3\text{r}^6}\Big)$
$=\text{a}^3(\text{r}^6+\text{r}^3+1)$
$=\text{a}^3+\text{a}^3\text{r}^3+\text{a}^3\text{r}^6$
$=\text{a}^3+(\text{ar})^3+(\text{ar}^2)^3$
$=\text{a}^3+\text{b}^3+\text{c}^3$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$

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