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Geometric Progressions — Maths STD 11 Science — Question

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

$a, b, c$ are in G.P.$a, b = ar, c = ar^2$
$\text{L.H.S}=\frac{\text{a}+\text{b}+\text{c}}{\text{a}^2+\text{b}^2+\text{c}^2}$
$=\frac{(\text{a}+\text{ar}+\text{ar}^2)^2}{\text{a}^2+\text{a}^2\text{r}^2+\text{a}^2\text{r}^4}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2(1+\text{r}^2+\text{r}^4)}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2\big[\big(1+2\text{r}^2+\text{r}^4\big)-\text{r}^2\big]}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2\big[\big(1+\text{r}^2+\text{r}\big)\big(1+\text{r}^2+\text{r}\big)\big]}$
$=\frac{\text{a}\big(1+\text{r}+\text{r}^2\big)}{\text{a}\big(1+\text{r}^2+\text{r}\big)}$
$=\frac{\text{a}+\text{ar}+\text{ar}^2}{\text{a}+\text{ar}^2-\text{ar}}$
$=\frac{\text{a}+\text{b}+\text{c}}{\text{a}-\text{b}+\text{c}}$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$

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