Question

If $x_1,x_2,x_3.........x_{4001}$ are in an 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 $30$

$x_1,x_2,x_3,....,x_{4001}$ are in AP

then, $\frac{1}{x_1},\frac{1}{x_2},\frac{1}{x_3},....,\frac{1}{x_{4001}}$ are in HP

Given $\frac{1}{x_1{x}_2}+\frac{1}{x_2{x}_3}+\frac{1}{x_3{x}_4}+.....+\frac{1}{x_{4000}{x}_{4001}}=10$

Dividing and multiplying by common difference $d$

$\frac{1}{x_1{x}_2}+\frac{1}{x_2{x}_3}+\frac{1}{x_3{x}_4}+.....+\frac{1}{x_{4000}{x}_{4001}}=10\frac{d}{d{x}_1{x}_2}+\frac{d}{d{x}_2{x}_3}+\frac{d}{d{x}_3{x}_4}+.....+\frac{d}{d{x}_{4000}{x}_{4001}}=10\frac{1}{d}[\frac{x_2-x_1}{x_1{x}_2}+\frac{x_3-x_2}{x_2{x}_3}+\frac{x_4-x_3}{x_3{x}_4}+.....+\frac{x_{4001}-x_{4000}}{x_{4000}{x}_{4001}}]=10\frac{1}{d}[\frac{1}{x_1}-\frac{1}{x_2}+\frac{1}{x_2}-\frac{1}{x_3}+\frac{1}{x_3}-....+\frac{1}{x_{4000}}-\frac{1}{x_{4001}}]=10\frac{1}{d}[\frac{1}{x_1}-\frac{1}{x_{4001}}]=10\frac{1}{d}\frac{x_{4001}-x_1}{x_{4001}{x}_1}=10\left(1\right)$

$a_n=a+(n-1)d$ $n^{th}$ term formula

$x_{4001}=x_1+(4001-1)d\frac{x_{4001}-x_1}{4000}=d\left(2\right)$

From $\left(1\right),\left(2\right)$

$\frac{d\times 4000}{d}=10\left(x_{4001}{x}_1\right)400=x_{4001}{x}_1(x_1-x_{4001}{)}^2=(x_1+x_{4001}{)}^2-4x_1{x}_{4001}(x_1-x_{4001}{)}^2=(50{)}^2-4\times 400(x_1-x_{4001}{)}^2=2500-1600=900(x_1-x_{4001})=\pm \sqrt{900}=\pm 30|x_1-x_{4001}|=30$