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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 . . . . . . .

Similar Questions

The sum of an infinite geometric progression is $\frac{4}{3}$ and the first term is $\frac{3}{4}$. The common ratio is

If the $4^{th}$,$7^{th}$,and $10^{th}$ terms of a Geometric Progression $(GP)$ are $a, b,$ and $c$ respectively,then:

If the $p^{\text{th}}$,$q^{\text{th}}$,and $r^{\text{th}}$ terms of a $G.P.$ are $a$,$b$,and $c$ respectively,prove that $a^{q-r} b^{r-p} c^{p-q} = 1$.

If $x, y, z$ are in geometric progression and $a^x = b^y = c^z$,then . . . . . .

If ${x_1}, {x_2}, {x_3}$ as well as ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module