Find three numbers in G.P. whose sum is 38 and their product is 1… - Vidyadip

Question

Find three numbers in G.P. whose sum is $38$ and their product is $1728$.

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Answer

Let the three number be $a, ar, ar^2$ in G.P., where a is first teror and $r$ is the common ratio.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 $8, 12, 18$.

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