Question

Let $x_1, x_2, x_3, …, x_{20}$ be in geometric progression with $x_1 = 3$ and the common ratio 1/2. A new data is constructed replacing each $x_i$ by $(x_i – i)^2$. If $x̄$ is the mean of new data, then the greatest integer less than or equal to $x̄$ is _________

Answer 1

In given geometric progression,

First term a=$x_1$=3 and common ratio r=$\frac{1}{2}$

∴ $x_r=3\frac {1}{2}^{r-1}$

$∑x_r=\frac{31-\frac{1}{2}^{20}} {1-\frac{1}{2}}=61-\frac{1}{2^{20}}$

$\sum_{i=1}^{20} x_i-i^2=\sum_{i=1}^{20} x_i^2-2x_ii+i^2$

Now, $\sum_{i=1}^{20} x_i^2=\frac{91-\frac{1} {4}^{20}}{1-\frac{1}{4}}=121-\frac{1}{2^{40}}$

$\sum_{i=1}^{20} i^2=\frac {1}{6}×20×21×41=2870$

and $\sum_{i=1}^{20} x_ii=s=3+2·3\frac{1}{2}+3·3\frac{1}{2^2}+4·3\frac{1}{2^3}+.... AGP$

=$62-\frac{22}{2^{20}}$

∴ $\bar x=\frac{12-\frac{12}{2^{40}}+2870-122-{\frac{22}{2^{20}}}}{20}$

⇒ $\bar x=\frac{2858}{2}+\frac{-12}{2^{40}}+\frac{22}{2^{20}}×\frac{1}{20}$

⇒ $\bar x=142$