Let x_1, x_2, x_3, \ldots, x_20 be in geometric progression with x_1=3 and the common ratio \frac12 . A new data is constructed replacing each x_i by (x_i-i)^2 . If \barx is the mean of new data, then the greatest integer less than or equal to \barx is \underline\hspace2cm .

Statistics PYQ — JEE Main 2022 · 28 Jul · Shift 1 · Q44 with Solution

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Correct answer: 142

Write the GP explicitly

Given: $x_1=3, \quad r=\frac12$ So, $x_i=3\left(\frac12\right)^{i-1}$

The new data are: $y_i=(x_i-i)^2$ Hence the mean of the new data is $\bar x=\frac1{20}\sum_{i=1}^{20}(x_i-i)^2$

Expand the square

$\bar x=\frac1{20}\sum_{i=1}^{20}\left(x_i^2-2ix_i+i^2\right)$ So, $\bar x=\frac1{20}\left(\sum_{i=1}^{20}x_i^2-2\sum_{i=1}^{20}ix_i+\sum_{i=1}^{20}i^2\right)$

We now compute these three sums separately.

Compute $\sum x_i^2$

Since $x_i=3\left(\frac12\right)^{i-1}$ we get $x_i^2=9\left(\frac14\right)^{i-1}$ Therefore, $\sum_{i=1}^{20}x_i^2=9\sum_{i=1}^{20}\left(\frac14\right)^{i-1}$ This is a GP with first term $1$ and ratio $\frac14$ : $\sum_{i=1}^{20}\left(\frac14\right)^{i-1}=\frac{1-(1/4)^{20}}{1-1/4}=\frac{1-(1/4)^{20}}{3/4}=\frac43\left(1-4^{-20}\right)$ Thus, $\sum_{i=1}^{20}x_i^2=9\cdot \frac43\left(1-4^{-20}\right)=12\left(1-4^{-20}\right)$

Compute $\sum ix_i$

$\sum_{i=1}^{20}ix_i=3\sum_{i=1}^{20} i\left(\frac12\right)^{i-1}$ Use the standard formula $\sum_{i=1}^{n} i r^{i-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}$ with $n=20$ , $r=\frac12$ :

=\frac{1-21\left(\frac12\right)^{20}+20\left(\frac12\right)^{21}}{(1/2)^2}$$ Now, $$20\left(\frac12\right)^{21}=10\left(\frac12\right)^{20}$$ so numerator becomes $$1-21\cdot 2^{-20}+10\cdot 2^{-20}=1-11\cdot 2^{-20}$$ Hence, $$\sum_{i=1}^{20} i\left(\frac12\right)^{i-1}=\frac{1-11\cdot 2^{-20}}{1/4}=4-44\cdot 2^{-20}$$ Therefore, $$\sum_{i=1}^{20}ix_i=3\left(4-44\cdot 2^{-20}\right)=12-132\cdot 2^{-20}$$ So, $$2\sum_{i=1}^{20}ix_i=24-264\cdot 2^{-20}$$ --- 5. **Compute $\sum i^2$** Using $$\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6}$$ for $n=20$: $$\sum_{i=1}^{20} i^2=\frac{20\cdot 21\cdot 41}{6}=2870$$ --- 6. **Substitute into the mean** $$\bar x=\frac1{20}\left[12\left(1-4^{-20}\right)-\left(24-264\cdot 2^{-20}\right)+2870\right]$$ Simplify: $$\bar x=\frac1{20}\left(12-12\cdot 4^{-20}-24+264\cdot 2^{-20}+2870\right)$$ $$\bar x=\frac1{20}\left(2858+264\cdot 2^{-20}-12\cdot 4^{-20}\right)$$ Since $$4^{-20}=2^{-40}$$ we have $$\bar x=\frac1{20}\left(2858+264\cdot 2^{-20}-12\cdot 2^{-40}\right)$$ Now, $$\frac{2858}{20}=142.9$$ and the remaining correction term is positive but very small: $$\frac{264\cdot 2^{-20}-12\cdot 2^{-40}}{20}\approx \frac{264}{20\cdot 1048576}>0$$ So, $$\bar x\approx 142.9000126$$ Therefore, $$\lfloor \bar x\rfloor = 142$$ --- 7. **Final answer** The greatest integer less than or equal to $\bar x$ is $$\boxed{142}$$

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View written solutionFree

Correct answer: 142

Write the GP explicitly

Given: $x_1=3, \quad r=\frac12$ So, $x_i=3\left(\frac12\right)^{i-1}$

The new data are: $y_i=(x_i-i)^2$ Hence the mean of the new data is $\bar x=\frac1{20}\sum_{i=1}^{20}(x_i-i)^2$

Expand the square

$\bar x=\frac1{20}\sum_{i=1}^{20}\left(x_i^2-2ix_i+i^2\right)$ So, $\bar x=\frac1{20}\left(\sum_{i=1}^{20}x_i^2-2\sum_{i=1}^{20}ix_i+\sum_{i=1}^{20}i^2\right)$

We now compute these three sums separately.

Compute $\sum x_i^2$

Compute $\sum ix_i$

$\sum_{i=1}^{20}ix_i=3\sum_{i=1}^{20} i\left(\frac12\right)^{i-1}$ Use the standard formula $\sum_{i=1}^{n} i r^{i-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}$ with $n=20$ , $r=\frac12$ :

=\frac{1-21\left(\frac12\right)^{20}+20\left(\frac12\right)^{21}}{(1/2)^2}$$ Now, $$20\left(\frac12\right)^{21}=10\left(\frac12\right)^{20}$$ so numerator becomes $$1-21\cdot 2^{-20}+10\cdot 2^{-20}=1-11\cdot 2^{-20}$$ Hence, $$\sum_{i=1}^{20} i\left(\frac12\right)^{i-1}=\frac{1-11\cdot 2^{-20}}{1/4}=4-44\cdot 2^{-20}$$ Therefore, $$\sum_{i=1}^{20}ix_i=3\left(4-44\cdot 2^{-20}\right)=12-132\cdot 2^{-20}$$ So, $$2\sum_{i=1}^{20}ix_i=24-264\cdot 2^{-20}$$ --- 5. **Compute $\sum i^2$** Using $$\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6}$$ for $n=20$: $$\sum_{i=1}^{20} i^2=\frac{20\cdot 21\cdot 41}{6}=2870$$ --- 6. **Substitute into the mean** $$\bar x=\frac1{20}\left[12\left(1-4^{-20}\right)-\left(24-264\cdot 2^{-20}\right)+2870\right]$$ Simplify: $$\bar x=\frac1{20}\left(12-12\cdot 4^{-20}-24+264\cdot 2^{-20}+2870\right)$$ $$\bar x=\frac1{20}\left(2858+264\cdot 2^{-20}-12\cdot 4^{-20}\right)$$ Since $$4^{-20}=2^{-40}$$ we have $$\bar x=\frac1{20}\left(2858+264\cdot 2^{-20}-12\cdot 2^{-40}\right)$$ Now, $$\frac{2858}{20}=142.9$$ and the remaining correction term is positive but very small: $$\frac{264\cdot 2^{-20}-12\cdot 2^{-40}}{20}\approx \frac{264}{20\cdot 1048576}>0$$ So, $$\bar x\approx 142.9000126$$ Therefore, $$\lfloor \bar x\rfloor = 142$$ --- 7. **Final answer** The greatest integer less than or equal to $\bar x$ is $$\boxed{142}$$

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