Question

If $x_1,x_2,x_3,....x_{4001}$ are in AP such that $\frac{1}{x_1{x}_2}+\frac{1}{x_2{x}_3}+...+\frac{1}{x_{4000}{x}_{4001}}=10$ and $x_2+x_{4000}=50$ then $|x_1-x_{4001}|=$

• A. $20$
• B. $30$
• C. $40$
• D. $100$

Answer 1

The correct option is B $30$

Let $x_1=a$ and common difference be $d$

$\therefore x_2=a+d,x_3=a+2d..........x_{4001}=a+4000d$

Given: $x_2+x_{4000}=50\Rightarrow a+d+a+3999d=50$

$a+4000d=50-a...\left(1\right)$

Also: $\frac{1}{x_1{x}_2}+\frac{1}{x_2{x}_3}+....+\frac{1}{x_{4000}{x}_{4001}}=10$

Multiply and divide by $d,$ we get,

$\frac{1}{d}(\frac{d}{a(a+d)}+\frac{d}{(a+d)(a+2d)}+....+\frac{d}{(a+3999d)(a+4000d)})=10$

$\frac{1}{d}((\frac{1}{a}-\frac{1}{a+d})+(\frac{1}{a+d}-\frac{1}{a+2d})+....+((\frac{1}{a+3999d}-\frac{1}{a+4000d}))=10$

$\frac{1}{d}(\frac{1}{a}-\frac{1}{a+4000d})=\mathrm{10.......}\left(2\right)$

$\frac{4000}{a(a+4000d)}=10$

Substitute value of $a+4000d$ by eq (1),

$\frac{4000}{a(50-a)}=10\Rightarrow 400=50a-a^2$

$a^2-50a+400=0\Rightarrow (a-40)(a-10)=0$

$a=10,40$

$\therefore a+4000d=50-a\to d=\frac{50-2a}{4000}$

$a=10\to d=\frac{30}{4000}=\frac{3}{400}$

$a=40\to d=-\frac{30}{4000}=-\frac{3}{400}$

$|x_1-x_{4001}|=|a-(a+4000d)|=|-4000d|=4000|d|$

$4000\times |\pm \frac{3}{400}|=30$