Geometric Progressions — MATHS STD 11 Science — Question

Then,

$\text{a} + \text{ar} +\text{ar}^2 = 38$

$\text{a} (1 + \text{r} + \text{r}^2) = 38\cdots(\text{i})$

and

$\text{(a)}\text{(ar)}\text{(ar)}^2=1728$

$\text{a}^3\text{r}^3=1728=4^33^3=(12)^3$

$\text{a}^3=\frac{12^3}{\text{r}^3}\Rightarrow\frac{12}{\text{r}}=\text{a}$

Putting $\text{a}=\frac{12}{\text{r}}\text{ in }(\text{i})$

$\frac{12}{\text{r}}(1 + \text{r} + \text{r}^2)=38$

$12+12\text{r}+12\text{r}^2=38\text{r}$

$12\text{r}^2-26\text{r}+12=0$

$6\text{r}^2-13\text{r}+6=0$

$6\text{r}^2-9\text{r}-4\text{r}+6=0$

$3\text{r}(3\text{r}-3)-2(3\text{r}-3)=0$

$\text{r}=\frac{3}{2},\frac{2}{3}$

$\text{a}=\frac{12}{\frac{3}{2}}=8\text{ or }\frac{12}{\frac23}=18$

$\therefore$ G.P. is 8, 12, 18.