Question

Let $x _1, x _2, x _3, \ldots ., x _{20}$ 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

$\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) $
$ =\displaystyle\sum_{ i =1}^{20}\left( x _{ i - i }\right)^2 $
$=\displaystyle\sum_{ i =1}^{20}\left( x _{ i }\right)^2+( i )^2-2 x _{ 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) $
$ \displaystyle\sum_{ i =1}^{20} i ^2=\frac{1}{6} \times 20 \times 21 \times 41=2870$
$\displaystyle\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}+\ldots \ldots . 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} $
$ \overline{ x } =\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right) \times \frac{1}{20} \\ {[\overline{ x }] } =142$