The sum of the first three terms of a G.P.

Question:

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

  1. $[-3, \infty)$

  2. $(-\infty, 9]$

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

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


Correct Option: , 4

Solution:

Let three terms of G.P. are $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}$, ar

product $=27$

$\Rightarrow a^{3}=27 \Rightarrow a=3$

$\mathrm{S}=\frac{3}{\mathrm{r}}+3 \mathrm{r}+3$

For $r>0$

$\frac{\frac{3}{r}+3 r}{2} \geq \sqrt{3^{2}} \quad($ By $\mathrm{AM} \geq \mathrm{GM})$

$\Rightarrow \frac{3}{r}+3 r \geq 6$........(1)

For $r<0 \quad \frac{3}{r}+3 r \leq-6$ $\ldots(2)$

From (1) & (2)

$S \in(-\infty-3] \cup[9, \infty]$

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