# A nice sum

If a) x1 + x2 + ...................+x1007 =(2014)2
b) x1 /(x1 +1)=x2 /(x2 +3)=.....................=x1007 /(x1007 +2013) then
x253 =?

1708
man111 singh ·

\hspace{-20}$Given$\bf{x_{1}+x_{2}+x_{3}+............+x_{1007}=(2014)^2..........(1)}$\\\\\\ And$\bf{\frac{x_{1}}{x_{1}+1}=\frac{x_{2}}{x_{2}+3}=\frac{x_{3}}{x_{3}+5}=...........=\frac{x_{1007}}{x_{1007}+2013}}$\\\\\\ Now Taking Reciprocal ..........\\\\\\$\bf{\Rightarrow \frac{x_{1}+1}{x_{1}}=\frac{x_{2}+3}{x_{2}}=\frac{x_{3}+5}{x_{3}}=...........=\frac{x_{1007}+2013}{x_{1007}}}$\\\\\\ Now Sub.$\bf{1}$from each Term, we get......\\\\\\$\bf{\Rightarrow \frac{1}{x_{1}}=\frac{3}{x_{2}}=\frac{5}{x_{3}}=.............=\frac{2013}{x_{1007}}}$\\\\\\ Now again Taking Reciprocal......\\\\\\ Let$\bf{\frac{x_{1}}{1}=\frac{x_{2}}{3}=\frac{x_{3}}{5}=............=\frac{x_{1007}}{2013}=k.............(2)}$\\\\\\ So$\bf{x_{1}+x_{2}+x_{3}+..........+x_{1007}=k\cdot \left(1+3+5+......+1007\right)=(1007)^2}$\\\\\\ Now from equation......$\bf{(1)}$\\\\\\$\bf{\Rightarrow (2014)^2=(1007)\Rightarrow k = 4}$\\\\\\ So from equation..........\bf{(2)} \\\\\\ \hspace{-20}\bf{\frac{x_{1}}{1}=\frac{x_{2}}{3}=\frac{x_{3}}{5}=.....=\frac{x_{n}}{(2n-1)}.......=\frac{x_{1007}}{2013}=k}$\\\\\\ So $\bf{\frac{x_{n}}{(2n-1)}=k\Rightarrow x_{n}=k\cdot (2n-1)}$\\\\\\ Now Put $\bf{n=253\;,k=4\;,}$ we get\\\\\\ $\bf{\Rightarrow x_{253}=4\cdot (2\times 253-1)=4\cdot 505=2020}$

466
Himanshu Giria ·

Xn=(2n-1)Xn-1

263
Sushovan Halder ·

The general term is xr/(xr +2r-1).Yours is also right.

• Himanshu Giria How to proceed after this...
263
Sushovan Halder ·

equate Tr=Tr+1.you will get some relation among all the coefficients

2305
Shaswata Roy ·

Is it 2020?

187
Swastik Haldar ·

2020.

591
Akshay Ginodia ·

Is the answer smthng close to 1007

• Sushovan Halder No. HINT:First calculate the general term in b).