Let x_1,x_2,,x_20 be in geometric progression with x_1=3 and common ratio 12.

Question

Let $x_1,x_2,\ldots,x_{20}$ be in geometric progression with $x_1=3$ and common ratio $\dfrac12$. A new data set is constructed by replacing each $x_i$ with $(x_i-i)^2$. If $\bar x$ is the mean of the new data, then the greatest integer $\le\bar x$ is \_\_\_\_\_\_\_\_.

Answer: 142

Accepted Answer

$$\bar x = \frac{1}{20}\sum_{i=1}^{20}\left(x_i^2-2ix_i+i^2\right).$$

Using $\sum x_i^2$ (a G.P. of ratio $1/4$), $\sum ix_i$ (arithmetico-geometric), and $\sum i^2$, the computation yields $\bar x \approx 142.6$, so $[\bar x]=142$.

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