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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 $.....$
$143$
$144$
$145$
$142$
Solution
$\sum x _{0}^{1}=\frac{3\left(1-\left(\frac{1}{2}\right)\right)^{20}}{1 \frac{-1}{2}}=6\left(1-\frac{1}{2^{20}}\right)$
$=\sum_{ i =1}^{20}\left( x _{ i – i }\right)^{2}$
$=\sum_{ i =1}^{20}\left( x _{ i }\right)^{2}+( i )^{2}-2 x _{ i } i$
Now $=\sum_{i=1}^{20}\left(x_{i}\right)^{2}=\frac{9\left(1-\left(\frac{1}{4}\right)\right)^{20}}{1-\frac{1}{4}}=12\left(1-\frac{1}{2^{40}}\right)$
$\sum_{i=1}^{20} i^{2}=\frac{1}{6} \times 20 \times 21 \times 41=2870$
$\sum_{ i =1}^{20} x _{ i } i = s =3+2.3 \frac{1}{2}+3.3 \frac{1}{2^{2}}+4.3 \frac{1}{2^{3}}+\ldots \ldots AGP$
$=6\left(2-\frac{22}{2^{20}}\right)$
$\overline{ x }=\frac{12-\frac{12}{2^{40}}+2870-12\left(2-\frac{22}{2^{20}}\right)}{20}$
$\overline{ x }=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right) \times \frac{1}{20}$
${[\overline{ x }]=142 }$
