1. Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a constant common ratio, $r$. The $i$-th term is given by $x_i = a \cdot r^{i-1}$, where $a$ is the first term.
  • Sum of the first $N$ terms of a GP: $S_N = \frac{a(1-r^N)}{1-r}$, for $r \neq 1$.
• Mean of a Data Set: For a data set $y_1, y_2, \ldots, y_N$, the mean is $\bar{y} = \frac{1}{N} \sum_{i=1}^{N} y_i$.
• Greatest Integer Function (Floor Function): $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$.

2. Step-by-Step Solution

Step 1: Define the terms of the Geometric Progression ($x_i$) The problem states that $x_1 = 3$ and the common ratio $r = \frac{1}{2}$. There are 20 terms. The general term of a GP is $x_i = x_1 \cdot r^{i-1}$. Substituting the given values, we get: $x_i = 3 \cdot \left(\frac{1}{2}\right)^{i-1}$ This expression defines each term of the original data set.

Step 2: Define the New Data and its Mean A new data set is constructed where each $x_i$ is replaced by $(x_i - 1)^2$. Let the terms of this new data set be $y_i = (x_i - 1)^2$. The mean of this new data set, denoted by $\bar{x}$, is given by the formula: $\bar{x} = \frac{1}{20} \sum_{i=1}^{20} y_i = \frac{1}{20} \sum_{i=1}^{20} (x_i - 1)^2$ Our goal is to evaluate this sum and then divide by 20.

Step 3: Expand the Summation Term Substitute the expression for $x_i$ into the sum and expand the squared term: $\sum_{i=1}^{20} (x_i - 1)^2 = \sum_{i=1}^{20} \left( 3 \left(\frac{1}{2}\right)^{i-1} - 1 \right)^2$ Using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$: $= \sum_{i=1}^{20} \left[ \left(3 \left(\frac{1}{2}\right)^{i-1}\right)^2 - 2 \cdot 3 \left(\frac{1}{2}\right)^{i-1} \cdot 1 + 1^2 \right]$ $= \sum_{i=1}^{20} \left[ 9 \left(\frac{1}{4}\right)^{i-1} - 6 \left(\frac{1}{2}\right)^{i-1} + 1 \right]$ We can now split this into three separate summations due to the linearity of summation: $= \sum_{i=1}^{20} 9 \left(\frac{1}{4}\right)^{i-1} - \sum_{i=1}^{20} 6 \left(\frac{1}{2}\right)^{i-1} + \sum_{i=1}^{20} 1$ Let's evaluate each of these three summations individually.

Step 4: Evaluate Each Summation Separately

• Sum A: $\sum_{i=1}^{20} 9 \left(\frac{1}{4}\right)^{i-1}$ This is a geometric progression with first term $a = 9 \cdot (1/4)^{1-1} = 9$, common ratio $R = \frac{1}{4}$, and number of terms $N = 20$. Using the GP sum formula $S_N = \frac{a(1-R^N)}{1-R}$: $S_A = \frac{9 \left(1 - \left(\frac{1}{4}\right)^{20}\right)}{1 - \frac{1}{4}} = \frac{9 \left(1 - \left(\frac{1}{4}\right)^{20}\right)}{\frac{3}{4}} = 12 \left(1 - \left(\frac{1}{4}\right)^{20}\right)$ Since $\left(\frac{1}{4}\right)^{20}$ is an extremely small positive number (approximately $9 \times 10^{-13}$), it can be approximated as 0 for calculations where the integer part is sought. $S_A \approx 12(1 - 0) = 12$

• Sum B: $\sum_{i=1}^{20} 6 \left(\frac{1}{2}\right)^{i-1}$ This is a geometric progression with first term $a = 6 \cdot (1/2)^{1-1} = 6$, common ratio $R = \frac{1}{2}$, and number of terms $N = 20$. Using the GP sum formula: $S_B = \frac{6 \left(1 - \left(\frac{1}{2}\right)^{20}\right)}{1 - \frac{1}{2}} = \frac{6 \left(1 - \left(\frac{1}{2}\right)^{20}\right)}{\frac{1}{2}} = 12 \left(1 - \left(\frac{1}{2}\right)^{20}\right)$ Similar to Sum A, $\left(\frac{1}{2}\right)^{20}$ is an extremely small positive number (approximately $9.5 \times 10^{-7}$). $S_B \approx 12(1 - 0) = 12$

• Sum C: $\sum_{i=1}^{20} 1$ This is simply the sum of 1, twenty times. $S_C = 1 \times 20 = 20$

Step 5: Combine the Sums to Find the Total Sum The total sum is $S_A - S_B + S_C$: $\sum_{i=1}^{20} (x_i - 1)^2 = S_A - S_B + S_C$ Substituting the precise expressions: $= \left[12 \left(1 - \left(\frac{1}{4}\right)^{20}\right)\right] - \left[12 \left(1 - \left(\frac{1}{2}\right)^{20}\right)\right] + 20$ $= 12 - 12\left(\frac{1}{4}\right)^{20} - 12 + 12\left(\frac{1}{2}\right)^{20} + 20$ $= 20 - 12\left(\frac{1}{2^{2}}\right)^{20} + 12\left(\frac{1}{2}\right)^{20}$ $= 20 - 12\left(\frac{1}{2}\right)^{40} + 12\left(\frac{1}{2}\right)^{20}$ The terms involving powers of $(1/2)$ are very small: $12\left(\frac{1}{2}\right)^{40} \approx 0$ $12\left(\frac{1}{2}\right)^{20} = \frac{12}{1048576} \approx 0.0000114$ So, the total sum is approximately $20 - 0 + 0.0000114 = 20.0000114$. $\text{Total Sum} \approx 20.0000114$

Step 6: Calculate the Mean ($\bar{x}$) The mean is the total sum divided by the number of terms (20): $\bar{x} = \frac{\text{Total Sum}}{20} = \frac{20.0000114}{20} = 1.00000057$

Step 7: Find the Greatest Integer Less Than or Equal to $\bar{x}$ The greatest integer less than or equal to $\bar{x}$ is denoted by $\lfloor \bar{x} \rfloor$. $\lfloor \bar{x} \rfloor = \lfloor 1.00000057 \rfloor = 1$

3. Common Mistakes & Tips

• Approximation of small terms: In JEE, when dealing with very small terms like $(1/2)^{20}$ or $(1/4)^{20}$ in a summation that is expected to result in an integer or a value close to an integer, it's generally safe to approximate these terms as 0 if they don't affect the integer part. However, always be mindful if the problem explicitly asks for high precision or if the answer options are very close. In this case, even precise calculation of these small terms shows they do not change the floor value.
• Expansion Errors: A common mistake is to incorrectly expand $(a-b)^2$. Double-check signs and coefficients.
• Double Check Formulas: Always verify the formulas for sum of GP before applying them.

4. Summary This problem demonstrates how to systematically approach summations by breaking them into manageable parts. It combines concepts from geometric progressions with basic statistics (mean) and algebra (summation properties). The ability to correctly identify and apply the appropriate summation formulas, along with careful algebraic manipulation, is crucial. The approximation of negligible terms is a practical skill for time-constrained exams. By carefully expanding the squared term and evaluating the resulting geometric series, we find the total sum, which leads to a mean slightly greater than 1, and thus a greatest integer of 1.

5. Final Answer The final answer is $\boxed{1}$.