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

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Question

Let three terms of G.P. are $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar}$ product $=27$

$\Rightarrow \mathrm{a}^{3}=27 \Rightarrow \math

Vedclass · Accepted Answer

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

Vedclass · Answer

Let three terms of G.P. are $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar}$ product $=27$

$\Rightarrow \mathrm{a}^{3}=27 \Rightarrow \mathrm{a}=3$

$S=\frac{3}{r}+3 r+3$

For ${r}>0$

$\frac{\frac{3}{r}+3 r}{2} \geq \sqrt{3^{2}} \quad($ By $A M \geq G M)$

$\Rightarrow \frac{3}{r}+3 r \geq 6$

For $r<0 \quad \frac{3}{r}+3 r \leq-6 \quad \ldots(2)$

From ( 1)  (2)

$\mathrm{S} \in(-\infty-3) \cup[9, \infty]$

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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