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Geometric Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSGeometric Progressions4 Marks
Question
If a, b, c are in G.P., prove that: $\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$
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Answer

$a, b, c$ are in G.P. $a, b=a r, c=a r^2$
$\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$
$\frac{\text{a}^2+\text{a}(\text{ar})+\text{a}^2\text{r}^2}{(\text{ar})\big(\text{ar}^2\big)+\big(\text{ar}^2\big)\text{a}+\text{a}(\text{ar})}=\frac{\text{ar}+\text{a}}{\text{ar}^2+\text{ar}}$
$\frac{\text{a}^2\big(1+\text{r}+\text{r}^2\big)}{\text{a}^2(\text{r}^3+\text{r}^2+\text{r})}=\frac{1+\text{r}}{\text{r}(1+\text{r})}$ $\frac{1}{\text{r}}=\frac{1}{\text{r}}$
$\text{L.H.S}=\text{R.H.S}$ So, $\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$

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