Question

The sum of the squares of three distinct real number which are in G.P. is S². If their sum is a S then a² belongs to –

• A. $\left( \frac{1}{3},1 \right) \cup (1,3)$
• B. $\left( \frac{1}{3},1 \right) \cup (3,\infty)$
• C. $\left( –\infty,\frac{1}{3} \right) \cup (3,\infty)$
• D. None of these

Answer 1

Answer: (A)

$\left( \frac{1}{3},1 \right) \cup (1,3)$

Answer 2

Let$\frac{a}{r}$, a , ar, three terms in G.P.

given a² $\left( \frac{1}{r^{2}} + 1 + r^{2} \right)$= s² ….(1)

and a $\left( \frac{1}{r} + r + 1 \right)$ = a s ….(2)

divide (1) by (2)

$\frac{s}{\alpha}$=$\frac{a\left( \frac{1}{r} + r - 1 \right)\left( \frac{1}{r} + r + 1 \right)}{\left( \frac{1}{r} + r + 1 \right)}$ = $a\left( \frac{1}{r} + r - 1 \right)$ ….(3)

from (2) & (3)

a = $\left( \frac{\alpha^{2} - 1}{2\alpha} \right)s$

put in equation (1)

$\left\lbrack \frac{1}{r^{2}} + 1 + r^{2} \right\rbrack = s^{2}$

Ž $\left( \frac{1}{r} - r \right)^{2} + 3$ = $\frac{4\alpha^{2}s^{2}}{(\alpha^{2} - 1)^{2}s^{2}}$

Ž $\frac{4\alpha^{2}}{(\alpha^{2} - 1)^{2}}$> 3 Ž 3a⁴ – 10 a ² + 3 < 0

Ž (3 a ² – 1) (a ² – 3) < 0

Ž $\frac{1}{3}$ < a ² < 3 , [a ² ¹ 1 Q we get a = 0]

a ² Ī $\left( \frac{1}{3},1 \right)$Č (1, 3).