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

<p>${x_1},{x_2},{x_3},\,.....,\,{x_{20}}$ are in G.P.</p> <p>${x_1} = 3,\,r = {1 \over 2}$</p> <p>$\overline x = {{\sum {x_i^2 - 2{x_i}i + {i^2}} } \over {20}}$</p> <p>$$ = {1 \over {20}}\left[ {12\left( {1 - {1 \over {{2^{40}}}}} \right) - 6\left( {4 - {{11} \over {{2^{18}}}}} \right) + 70 \times 41} \right]$$</p> <p>$$\left\{ {\matrix{ {S = 1 + 2\,.\,{1 \over 2} + 3\,.\,{1 \over {{2^2}}}\, + \,....} \cr {{S \over 2} = {1 \over 2} + {2 \over {{2^2}}}\, + \,....} \cr } } \right.$$</p> <p>$$\left. {{S \over 2} = 2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}} = 4 - {{11} \over {{2^{18}}}}} \right\}$$</p> <p>$\therefore$ $$[\overline x ] = \left[ {{{2858} \over {20}} - \left( {{{12} \over {240}} - {{66} \over {{2^{18}}}}} \right)\,.\,{1 \over {20}}} \right]$$</p> <p>$= 142$</p>