JEE Main 2022 — Mathematics Algebra

Given,

${x}_{1},{x}_{2},{x}_{3},\ldots ..,{x}_{20}$x1​,x2​,x3​,…..,x20​ are in G.P with common difference $\frac{1}{2}$21​ and first term as $3$3

Now using the formula for sum of G.P we get,

$$\sum _{i=1}^{20}{x}_{i}=\frac{3(1-{(\frac{1}{2})}^{20})}{1-\frac{1}{2}}=6(1-\frac{1}{{2}^{20}})$$i=1∑20​xi​=1−21​3(1−(21​)20)​=6(1−2201​)

Now new sum of G.P with new data will be$\displaystyle =\sum _{i=1}^{20}{({x}_{i}-i)}^{2}$=i=1∑20​(xi​−i)2

$$=\sum _{i=1}^{20}{({x}_{i})}^{2}+{(i)}^{2}-2{x}_{i}i$$=i=1∑20​(xi​)2+(i)2−2xi​i

Now $\displaystyle \sum _{i=1}^{20}{({x}_{i})}^{2}=\frac{9(1-{(\frac{1}{4})}^{20})}{1-\frac{1}{4}}=12(1-\frac{1}{{2}^{40}})$i=1∑20​(xi​)2=1−41​9(1−(41​)20)​=12(1−2401​)

And $\displaystyle \sum _{i=1}^{20}{i}^{2}=\frac{1}{6}\times 20\times 21\times 41=2870$i=1∑20​i2=61​×20×21×41=2870

And $\displaystyle \sum _{i=1}^{20}{x}_{i}i=3+2.3\frac{1}{2}+3.3\frac{1}{{2}^{2}}+4.3\frac{1}{{2}^{3}}+\ldots \ldots .\mathrm{AGP}$i=1∑20​xi​i=3+2.321​+3.3221​+4.3231​+…….AGP

$$=6(4-\frac{11}{{2}^{18}})$$=6(4−21811​)

So, mean of the new data will be,

$$x=\frac{12-\frac{12}{{2}^{40}}+2870-6(4-\frac{11}{{2}^{18}})}{20}$$x=2012−24012​+2870−6(4−21811​)​

$$\Rightarrow x=\frac{2858}{20}+(\frac{-12}{{2}^{40}}+\frac{66}{{2}^{18}})\times \frac{1}{20}$$⇒x=202858​+(240−12​+21866​)×201​

Now $\Rightarrow [x]=[142.4+0.00001]=142$⇒[x]=[142.4+0.00001]=142