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 $$\text{R-(a,b)}$$, then $$a^{2}+b^{2}$$ is equal to______

Let the first three terms of the GP be $$\dfrac{a}{r},a,ar$$

Product of the first three terms = 27

$$\dfrac{a}{r}\times a\times ar=27$$

$$a=3$$

Sum of the first three terms $$=\dfrac{3}{r}+3+3r=3+3\left(r+\dfrac{1}{r}\right)$$

Now if $$r>0$$, then $$r+\dfrac{1}{r}\ge2$$ => $$3\left(r+\dfrac{1}{r}\right)\ge6$$ => $$3+3\left(r+\dfrac{1}{r}\right)\ge9$$

Now if $$r<0$$, then $$r+\dfrac{1}{r}\le-2$$ => $$3\left(r+\dfrac{1}{r}\right)\le-6$$ => $$3+3\left(r+\dfrac{1}{r}\right)\le-3$$

Therefore, the possible values of the sum of the first three terms will be every real number not in the range of $$\left(-3,9\right)$$

Thus, the values will be $$R-\left(-3,9\right)$$

$$a=-3$$ and $$b=9$$

$$a^2+b^2=(-3)^2+9^2=90$$