The roots of the equation $${\left| {x - 1} \right|^2} - 4\left| {x - 1} \right| + 3 = 0$$
A.
form an A.P.
B.
form a G.P.
C.
form an H.P.
D.
do not form any progression
Answer :
form an A.P.
Solution :
The given eq. can be written as
$$\eqalign{
& {\left| {x - 1} \right|^2} - 4\left| {x - 1} \right| + 3 = 0 \cr
& \Rightarrow \left( {\left| {x - 1} \right| - 3} \right)\left( {\left| {x - 1} \right| - 1} \right) = 0 \cr
& {\text{If }}\left| {x - 1} \right| - 3 = 0 \cr
& \Rightarrow x - 1 = \pm 3 \cr
& \Rightarrow x = - 2\,\,{\text{or 4}} \cr
& {\text{If }}\left| {x - 1} \right| - 1 = 0 \cr
& \Rightarrow x - 1 = \pm 1 \cr
& \Rightarrow x = 0\,\,{\text{or 2}} \cr} $$
The four roots are $$– 2, 0, 2, 4$$ and are in A.P.
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-