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

• A. $(-\infty ,-9]\cup [3,\infty )$
• B. $[-3,\infty )$
• C. $(-\infty ,-9]$
• D. $(-\infty ,-3]\cup [9,\infty )$

Answer 1

The correct option is D

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

Explanation for the correct answer:

Let the three terms of the geometric progression be $\frac{a}{r},a,ar$

The product of these term is $27$

$\Rightarrow$ $\frac{a}{r}\times a\times ar=27$

$\Rightarrow$ $a^3=27$

$\Rightarrow$ $a=3$

Hence, the geometric progression is $\frac{3}{r},3,3r$

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

For $r>0$

$\frac{{\displaystyle \frac{3}{r}}+3r}{2}\ge \sqrt{\frac{3}{r}\times 3r}$ …[AM-GM Inequality]

$\Rightarrow$ $\frac{3}{r}+3r\ge 6$

$\Rightarrow$$\frac{3}{r}+3r+3\ge 9$

$\Rightarrow$ $S\ge 9$

$\Rightarrow$ $S\in [9,\infty )$ $...\left(i\right)$

For $r<0$

$\Rightarrow$ $\frac{3}{r}+3r\le -6$

$\Rightarrow$$\frac{3}{r}+3r+3\le -3$

$\Rightarrow$ $S\le -3$

$\Rightarrow$ $S\in (-\infty ,-3]$ $...\left(ii\right)$

From $\left(i\right),\left(ii\right)$ we get

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

Hence, option $\left(D\right)$ is the correct answer