JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 22)

Let $$x_{1}, x_{2}, x_{3}, \ldots, x_{20}$$ be in geometric progression with $$x_{1}=3$$ and the common ratio $$\frac{1}{2}$$. 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
142

Explanation

$${x_1},{x_2},{x_3},\,.....,\,{x_{20}}$$ are in G.P.

$${x_1} = 3,\,r = {1 \over 2}$$

$$\overline x = {{\sum {x_i^2 - 2{x_i}i + {i^2}} } \over {20}}$$

$$ = {1 \over {20}}\left[ {12\left( {1 - {1 \over {{2^{40}}}}} \right) - 6\left( {4 - {{11} \over {{2^{18}}}}} \right) + 70 \times 41} \right]$$

$$\left\{ {\matrix{ {S = 1 + 2\,.\,{1 \over 2} + 3\,.\,{1 \over {{2^2}}}\, + \,....} \cr {{S \over 2} = {1 \over 2} + {2 \over {{2^2}}}\, + \,....} \cr } } \right.$$

$$\left. {{S \over 2} = 2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}} = 4 - {{11} \over {{2^{18}}}}} \right\}$$

$$\therefore$$ $$[\overline x ] = \left[ {{{2858} \over {20}} - \left( {{{12} \over {240}} - {{66} \over {{2^{18}}}}} \right)\,.\,{1 \over {20}}} \right]$$

$$ = 142$$

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