Question
Find three numbers in G.P. whose sum is 38 and their product is 1728.
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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| C1 | | C2 | |
| | Probability | | Written Description. |
| a. | 0.95 | i. | An incorrect assignment. |
| b. | 0.02 | ii. | No chance of happening. |
| c. | -0.3 | iii. | As much chance of happening as not. |
| d. | 0.5 | iv. | Very likely to happen. |
| e. | 0 | v. | Very little chance of happening. |
$(\text{x}+\text{y})+(\text{x}^2+\text{xy}+\text{y}^2)+(\text{x}^3+\text{x}^2\text{y}+\text{xy}^2+\text{y})+\ ...\text{ to n terms;}$