If the sum of the first 100 terms of an AP is -1 and the sum of even terms lying in first 100 terms is1. Then

$\begin{array}{l}{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+\dots \dots \dots +{\mathrm{x}}_{100}=\frac{100}{2}\left({\mathrm{x}}_1+{\mathrm{x}}_{100}\right)=-1\\ {\mathrm{x}}_2+{\mathrm{x}}_4+\dots \dots +{\mathrm{x}}_{100}=\frac{50}{2}\left({\mathrm{x}}_1+\mathrm{d}+{\mathrm{x}}_{100}\right)=1\end{array}$
 $$\begin{gathered} \mathrm{from} \mathrm{these} \\ \mathrm{d}=\frac{3}{50} \\ from \left(1\right){\mathrm{x}}_1+{\mathrm{x}}_1+99\mathrm{d}=\frac{-1}{50} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\mathrm{x}}_1=\frac{-149}{50} \\ {100}^{th }term=x_1+99d \\ = \frac{-149}{50}+99(\frac{3}{50} ) \\ =\frac{-149+297}{50} \\ =\frac{148}{50} \\ =\frac{74}{25} \end{gathered}$$