a, b = ar, c = ar2
$\text{L.H.S}=\frac{\text{a}+\text{b}+\text{c}}{\text{a}^2+\text{b}^2+\text{c}^2}$
$=\frac{(\text{a}+\text{ar}+\text{ar}^2)^2}{\text{a}^2+\text{a}^2\text{r}^2+\text{a}^2\text{r}^4}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2(1+\text{r}^2+\text{r}^4)}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2\big[\big(1+2\text{r}^2+\text{r}^4\big)-\text{r}^2\big]}$
$=\frac{\text{a}^2(1+\text{r}+\text{r}^2)^2}{\text{a}^2\big[\big(1+\text{r}^2+\text{r}\big)\big(1+\text{r}^2+\text{r}\big)\big]}$
$=\frac{\text{a}\big(1+\text{r}+\text{r}^2\big)}{\text{a}\big(1+\text{r}^2+\text{r}\big)}$
$=\frac{\text{a}+\text{ar}+\text{ar}^2}{\text{a}+\text{ar}^2-\text{ar}}$
$=\frac{\text{a}+\text{b}+\text{c}}{\text{a}-\text{b}+\text{c}}$
$=\text{R.H.S}$
$\therefore\text{R.H.S}=\text{L.H.S}$
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Find the middle terms(s) in the expansion of:
$\Big(3-\frac{\text{x}^{3}}{6}\Big)^{7}$
| xi | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
| fi | 3 | 2 | 3 | 2 | 6 | 3 | 3 |