Question

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 $\bar{x}$ is the mean of new data, then the greatest integer less than or equal to $\bar{x}$ is __

Answer 1

Answer: 142

${\sum }_{x_0^1}^1=\frac{3{\left(1-\left(\frac{1}{2}\right)\right)}^{20}}{1\frac{-1}{2}}=6\left(1-\frac{1}{2^{20}}\right)$
$=\sum _{i=1}^{20}{\left(x_{i-i}\right)}^2$
$=\sum _{i=1}^{20}{\left(x_i\right)}^2+(i{)}^2-2x_{i}i$
$\text{ Now }={\sum }_{i=1}^{20}{\left(x_i\right)}^2=\frac{9{\left(1-\left(\frac{1}{4}\right)\right)}^{20}}{1-\frac{1}{4}}=12\left(1-\frac{1}{2^{40}}\right)$
$\sum _{i=1}^{20}{i}^2=\frac{1}{6}\times 20\times 21\times 41=2870$
$\sum _{i=1}^{20}{x}_{i}i=s=3+2.3\frac{1}{2}+3.3\frac{1}{2^2}+4.3\frac{1}{2^3}+\dots \dots .AGP$
$=6\left(2-\frac{22}{2^{20}}\right)$
$\overline{x}=\frac{12-\frac{12}{2^{40}}+2870-12\left(2-\frac{22}{2^{20}}\right)}{20}$
$$\begin{gathered} \overline{x}=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right)\times \frac{1}{20} \\ [\overline{x}]=142 \end{gathered}$$