Verified Solution • 2025–26 Session moderate

QJEEM23Ma715a: JEE Main 2023 Mathematics - Class 11 JEEM23Ma715a Solution: Let be in an arithmetic progression, with and their mean equal…

Let x1,x2,…,x100x_{1}, x_{2}, \ldots, x_{100} be in an arithmetic progression, with x1=2x_{1}=2 and their mean equal to 200 . If yi=i(xi−i),1≤i≤100y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100 , then the mean of y1,y2,…,y100y_{1}, y_{2}, \ldots, y_{100} is :

Multiple Choice Options

  • A 10051.5
  • B 10049.5
  • C 10100
  • D 10101.5

Step-by-Step Verified Solution

We have, mean of x1,x2……x100=200x_1, x_2 \ldots \ldots x_{100}=200

Where, x1,x2…x100x_1, x_2 \ldots x_{100} are in AP with first term as 2.

 Mean amp;=200amp;=∑i=1100xi100=200\begin{aligned}\text { Mean } & =200 \\\\& =\frac{\sum\limits_{i=1}^{100} x_i}{100}=200\end{aligned}

1002×[2×2+99d]=20000⇒4+99d=400⇒99d=396d=4\begin{aligned}\frac{100}{2} \times[2 \times 2+99 d] =20000 \\\\\Rightarrow 4+99 d =400 \\\\\Rightarrow 99 d =396 \\\\d =4\end{aligned}

Also,

yiamp;=i(xi−i)amp;=i[2+(i−1)4−i]amp;=i[3i−2]amp;=3i2−2i\begin{aligned}y_i & =i\left(x_i-i\right) \\\\& =i[2+(i-1) 4-i] \\\\& =i[3 i-2] \\\\& =3 i^2-2 i\end{aligned}

 Required mean amp;=∑i=1100yi100amp;=1100[∑i=1100(3i2−2i)]amp;=1100[3×100×101×2016−2×100×1012]amp;=203012−101amp;=10049.50\begin{aligned}\text { Required mean } & =\frac{\sum\limits_{i=1}^{100} y_i}{100} \\\\& =\frac{1}{100}\left[\sum_{i=1}^{100}\left(3 i^2-2 i\right)\right] \\\\& =\frac{1}{100}\left[\frac{3 \times 100 \times 101 \times 201}{6}-2 \times \frac{100 \times 101}{2}\right]\\\\& =\frac{20301}{2}-101 \\\\& =10049.50\end{aligned}

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