Question

Given the sequence of numbers $x_1,x_2,x_3,........x_{2013}$ which satisfies $\frac{x_1}{x_1+1}=\frac{x_2}{x_2+3}=\frac{x_3}{x_3+5}=......=\frac{x_{2013}}{x_{2013}+4025}$, nature of the sequence is

• A. $A.P$
• B. $G.P$
• C. $H.P$
• D. $A.G.P$

Answer 1

The correct option is A $A.P$

Given,

$\frac{x_1}{x_1+1}=\frac{x_2}{x_2+3}=\frac{x_3}{x_3+5}=........=\frac{x_{2013}}{x_{2013}+4025}$

$\Rightarrow \frac{x_1+1}{x_1}=\frac{x_2+3}{x_2}=\frac{x_3+5}{x_3}=........=\frac{x_{2013}+4025}{x_{2013}}$ (Taking reciprocal)

$\Rightarrow 1+\frac{1}{x_1}=1+\frac{3}{x_2}=1+\frac{5}{x_3}=........=1+\frac{4025}{x_{2013}}$

$\Rightarrow \frac{1}{x_1}=\frac{3}{x_2}=\frac{5}{x_3}=........=\frac{4025}{x_{2013}}$

$\Rightarrow \frac{x_1}{1}=\frac{x_2}{3}=\frac{x_3}{5}=........=\frac{x_{2013}}{4025}=k\left(say\right)$

Therefore, $x_1=k$, $x_2=3k$, $x_3=5k$,..... and so on.

$\Rightarrow x_1-x_2=x_2-x_3=x_3-x_4=.......=2k$.

Since the difference between two consecutive terms is constant($2k$), the given series is an A.P.

Hence, Option $A$ is correct.