Question

Let $A(a,b)$ and $B(0,0)$ be two fixed points. If $M_1$ is the mid point of line segment $AB$, $M_2$ is the mid point of line segment $A{M}_1$ and $M_3$ is the mid point of line segment $A{M}_2$ and so on, then which of the following is/are true?

• A. $M_3=(\frac{7a}{8},\frac{7b}{8})$
• B. $M_4=(\frac{a}{16},\frac{b}{16})$
• C. $M_5=(\frac{31a}{32},\frac{31b}{32})$
• D. $M_2=(\frac{3a}{2},\frac{3b}{2})$

Answer 1

Answer: (A), (C)

The correct options are
A $M_3=(\frac{7a}{8},\frac{7b}{8})$
C $M_5=(\frac{31a}{32},\frac{31b}{32})$

$A(a,b),B(0,0)\Rightarrow M_1=(\frac{a}{2},\frac{b}{2})$
$A(a,b)\text{and}{M}_1=(\frac{a}{2},\frac{b}{2})\Rightarrow M_2=(\frac{3a}{4},\frac{3b}{4})$
$A(a,b)\text{and}{M}_2=(\frac{3a}{4},\frac{3b}{4})$
$\Rightarrow M_3=(\frac{7a}{8},\frac{7b}{8})$
$A(a,b)\text{and}{M}_3=(\frac{7a}{8},\frac{7b}{8})\Rightarrow M_4=(\frac{15a}{16},\frac{15b}{16})...$
We can observe the coefficient of $a$ or $b$ are $\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\dots$
Which can be written as
$1-{\left(\frac{1}{2}\right)}^1,1-{\left(\frac{1}{2}\right)}^2,1-{\left(\frac{1}{2}\right)}^3,1-{\left(\frac{1}{2}\right)}^4,\dots$
$\therefore \text{In}{M}_n$ the coefficient of $a$ or $b$ is $1-{\left(\frac{1}{2}\right)}^n$

Hence, $M_3=(\frac{7a}{8},\frac{7b}{8})$ and $M_5=(\frac{31a}{32},\frac{31b}{32})$