In a G.P., if the product of the first three | vedclass

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(B) Let the first three terms of the $G.P.$ be $\frac{A}{r}, A, Ar$. Given their product is $27$: $\frac{A}{r} \cdot A \cdot Ar = 27 \implies A^3 = 27

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Let the first three terms of the $G.P.$ be $\frac{A}{r}, A, Ar$. Given their product is $27$: $\frac{A}{r} \cdot A \cdot Ar = 27 \implies A^3 = 27 \implies A = 3$. The sum of the first three terms is $S = \frac{3}{r} + 3 + 3r = 3 \left( r + \frac{1}{r} + 1 ight)$. We know that for any real $r 
eq 0$, $r + \frac{1}{r} \geq 2$ or $r + \frac{1}{r} \leq -2$. If $r + \frac{1}{r} \geq 2$, then $S \geq 3(2 + 1) = 9$. If $r + \frac{1}{r} \leq -2$, then $S \leq 3(-2 + 1) = -3$. Thus, the set of possible values for $S$ is $(-\infty, -3] \cup [9, \infty)$, which is $\mathbb{R} - (-3, 9)$. Comparing this with $\mathbb{R} - (a, b)$, we get $a = -3$ and $b = 9$. Therefore, $a^2 + b^2 = (-3)^2 + 9^2 = 9 + 81 = 90$.

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 . . . . . . .

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