The sum of the first three terms of a G.P. is | vedclass

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(D) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that their product is $27$,we have $\frac{a}{r} \cdot a \cdot ar = 27$,which impli

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$(-\infty, -3] \cup [9, \infty)$

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(D) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that their product is $27$,we have $\frac{a}{r} \cdot a \cdot ar = 27$,which implies $a^3 = 27$,so $a = 3$.The sum of these terms is $S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + r + 1)$.Case $1$: If $r > 0$,by $AM \geq GM$,we have $\frac{r + \frac{1}{r}}{2} \geq \sqrt{r \cdot \frac{1}{r}} = 1$,so $r + \frac{1}{r} \geq 2$.Thus,$S = 3(r + \frac{1}{r} + 1) \geq 3(2 + 1) = 9$.Case $2$: If $r  0$. Then $r + \frac{1}{r} = -(k + \frac{1}{k}) \leq -2$.Thus,$S = 3(r + \frac{1}{r} + 1) \leq 3(-2 + 1) = -3$.Therefore,$S \in (-\infty, -3] \cup [9, \infty)$.

The sum of the first three terms of a $G.P.$ is $S$ and their product is $27$. Then all such $S$ lie in

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