In a G.P., if the product of the first three terms is 27 and the set of all possible value…

Question

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $\mathbb{R}-(a, b)$, then $a^2+b^2$ is equal to
$\_\_\_\_$ .

Answer: 90

Accepted Answer

Product of first three terms is 27.

Let the terms be $a/r,a,ar$.

$(a/r)(a)(ar)=27\Rightarrow {a}^{3}=27\Rightarrow a=3$

Sum $S=\frac{3}{r}+3+3r=3(r+\frac{1}{r}+1)$.

Find the range of the sum

The sum $S=3(\frac{1}{r}+1+r)$.

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality for $r+\frac{1}{r}$ :

If $r>0$, then $r+\frac{1}{r}\ge 2$, so $S\ge 3(2+1)=9$.

If $rThe possible values for the sum are $(-\infty,-3]\cup [9,\infty)$.

This is expressed as $R-(-3,9)$.

Thus, $a=-3$ and $b=9$.

The final value is $

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