Let $x_1,x_2,x_3,\dots ..,x_{20}$ be in geometric progression with $x_1=3$ and the common ration $\frac{1}{2}$. A new data is constructed replacing each $x_i$ 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

Given,

$x_1,x_2,x_3,\dots ..,x_{20}$ are in G.P with common difference $\frac{1}{2}$ and first term as $3$

Now using the formula for sum of G.P we get,

$\sum _{i=1}^{20}{x}_i=\frac{3\left(1-{\left(\frac{1}{2}\right)}^{20}\right)}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^{20}}\right)$

Now new sum of G.P with new data will be$=\sum _{i=1}^{20}{\left(x_i-i\right)}^2$

$=\sum _{i=1}^{20}{\left(x_i\right)}^2+{\left(i\right)}^2-2x_{i}i$

Now $\sum _{i=1}^{20}{\left(x_i\right)}^2=\frac{9\left(1-{\left(\frac{1}{4}\right)}^{20}\right)}{1-\frac{1}{4}}=12\left(1-\frac{1}{2^{40}}\right)$

And $\sum _{i=1}^{20}{i}^2=\frac{1}{6}\times 20\times 21\times 41=2870$

And $\sum _{i=1}^{20}{x}_{i}i=3+2.3\frac{1}{2}+3.3\frac{1}{2^2}+4.3\frac{1}{2^3}+\dots \dots .\mathrm{AGP}$

$=6\left(4-\frac{11}{2^{18}}\right)$

So, mean of the new data will be,

 $\overline{x}=\frac{12-\frac{12}{2^{40}}+2870-6\left(4-\frac{11}{2^{18}}\right)}{20}$

$\Rightarrow \overline{x}=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{66}{2^{18}}\right)\times \frac{1}{20}$

Now $\Rightarrow \left[\overline{x}\right]=\left[142.4+0.00001\right]=142$