Let $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots ., \mathrm{x}_{20}$ be in geometric progression with $\mathrm{x}_{1}=3$ and the common ration $\frac{1}{2}$. A new data is constructed replacing each $\mathrm{x}_{\mathrm{i}}$ by $\left(\mathrm{x}_{\mathrm{i}}-\mathrm{i}\right)^{2}$. If $\overline{\mathrm{x}}$ is the mean of new data, then the greatest integer less than or equal to $\bar{x}$ is ______.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Statistics

Let

x_{1}, x_{2}, x_{3}, \dots ., 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}

(x_{i} - i)^{2}

. If

\overline{x}

is the mean of new data, then the greatest integer less than or equal to

\overset{x}{ˉ}

is ______.

Answer: 142

The GP has

x_{i} = 3 \cdot (\frac{1}{2})^{i - 1}

.

The new data consists of

(x_{i} - i)^{2} = x_{i}^{2} + i^{2} - 2 x_{i} i

.

Compute:

\sum_{i = 1}^{20} x_{i}^{2} = \frac{9 ( 1 - ( 1/4 ) ^{20} )}{1 - 1/4} = 12 (1 - \frac{1}{2 ^{40}})

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

.

For

\sum_{i = 1}^{20} x_{i} i

, this is an AGP (arithmetic-geometric progression) sum. Using the formula, we get

\sum x_{i} i = 6 (2 - \frac{22}{2 ^{20}})

.

Therefore,

\overset{x}{ˉ} = \frac{12 - \frac{12}{2 ^{40}} + 2870 - 12 ( 2 - \frac{22}{2 ^{20}} )}{20} = \frac{2858 + small terms}{20}

.

The dominant term gives

\overset{x}{ˉ} \approx \frac{2858}{20} = 142.9

.

Thus,

[\overset{x}{ˉ}] = 142

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer