Let $$f\left( x \right) = 2x + 1.$$ Then the number of real values of $$x$$ for which the three unequal numbers $$f\left( x \right),f\left( {2x} \right),f\left( {4x} \right)$$ are in G.P. is
A.
1
B.
2
C.
0
D.
none of these
Answer :
0
Solution :
$$2x + 1, 4x + 1, 8x +1$$ are in G.P.
$$ \Rightarrow \,\,{\left( {4x + 1} \right)^2} = \left( {2x + 1} \right)\left( {8x + 1} \right)$$
⇒ $$x = 0$$ and for this value $$f\left( x \right),f\left( {2x} \right),f\left( {4x} \right)$$ are equal.
Releted MCQ Question on Algebra >> Sequences and Series
Releted Question 1
If $$x, y$$ and $$z$$ are $${p^{{\text{th}}}},{q^{{\text{th}}}}\,{\text{and }}{r^{{\text{th}}}}$$ terms respectively of an A.P. and also of a G.P., then $${x^{y - z}}{y^{z - x}}{z^{x - y}}$$ is equal to:
If $$a, b, c$$ are in G.P., then the equations $$a{x^2} + 2bx + c = 0$$ and $$d{x^2} + 2ex + f = 0$$ have a common root if $$\frac{d}{a},\frac{e}{b},\frac{f}{c}$$ are in-