$$\begin{gathered} Let x_1, x_2, x_3........, x_{20} be in geometric progression \\ with x_1=3 and the common ratio\frac{1}{2}. A new data \\ is constructed replacing each x_1 by{\left(x_i-i\right)}^2. If \overline{x} is \\ the mean of new data.then the greatest integer less \\ than or equal to \overline{x} is\_\_\_\_\_\_\_\_\_\_\_. \end{gathered}$$

$$\begin{gathered} Let x_1, x_2, x_3........, x_{20} are in G.P with x_1=3,r=\frac{1}{2} \\ \therefore x_r=3{\left(\frac{1}{2}\right)}^{r-1} \\ \underset{ }{\sum x_r^{ }=\frac{3{\left(1-\left({\displaystyle \frac{1}{2}}\right)\right)}^{20}}{1{\displaystyle \frac{-1}{2}}}=6\left(1-\frac{1}{2^{20}}\right)} \\ =\sum _{i=1}^{20}{\left(x_i-i\right)}^2 \\ =\sum _{i=1}^{20}{\left(x_i\right)}^2+{\left(i\right)}^2-2x_{i}i \\ Now=\sum _{i=1}^{20}{\left(x_i\right)}^2=\frac{9{\left(1-\left({\displaystyle \frac{1}{4}}\right)\right)}^{20}}{1{\displaystyle -\frac{1}{4}}}=12\left(1-\frac{1}{2^{40}}\right) \\ \sum _{i=1}^{20}{i}^2=\frac{1}{6}\times 20\times 21\times 41=2870, \\ and \sum _{i=1}^{20}{x}_i i=s=3+2.3\frac{1}{2}+3.3\frac{1}{2^2}+4.3\frac{1}{2^3}+....... AGP \\ =6\left(2-\frac{22}{2^{20}}\right) \\ \therefore \overline{x}=\frac{12-{\displaystyle \frac{12}{2^{40}}}+2870-12\left(2-{\displaystyle \frac{22}{2^{20}}}\right)}{20} \\ \Rightarrow \overline{x}=\frac{2858}{2}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right)\times \frac{1}{20} \\ \Rightarrow \left[\overline{x}\right]=142 \end{gathered}$$