JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $x_{1}, x_{2}, \dots, x_{100}$ be in an arithmetic progression, with $x_{1} = 2$ and their mean equal to $200$ . If $y_{i} = i (x_{i} - i), 1 \leq i \leq 100$ , then the mean of $y_{1}, y_{2}, \dots, y_{100}$ is:

Choose the correct answer:

Explanation

1. $x_i$ sequence (AP) ko samajhna:

Maana ki AP ka pehla term $a = 2$ hai aur common difference $d$ hai.

Hume diya gaya hai ki 100 terms ka mean 200 hai.

\text{Mean} = \frac{\text{Sum}}{100} = 200 \implies \text{Sum} (S_{100}) = 20000

AP ke sum ka formula: $S_n = \frac{n}{2} [2a + (n-1)d]$

20000 = \frac{100}{2} [2(2) + (100-1)d]

20000 = 50 [4 + 99d]

400 = 4 + 99d \implies 99d = 396 \implies d = 4

Toh $x_i$ ka general term hoga:

x_i = a + (i-1)d = 2 + (i-1)4 = 4i - 2

2. $y_i$ ki value nikalna:

Sawal ke anusar, $y_i = i(x_i - i)$ hai.

y_i = i(4i - 2 - i)

y_i = i(3i - 2) = 3i^2 - 2i

3. $y_i$ ka mean nikalna:

Mean of $y = \frac{\sum_{i=1}^{100} y_i}{100} = \frac{\sum_{i=1}^{100} (3i^2 - 2i)}{100}$

\text{Sum of } y_i = 3 \sum i^2 - 2 \sum i

Hum jaante hain:

$\sum i = \frac{n(n+1)}{2} = \frac{100 \times 101}{2} = 5050$
$\sum i^2 = \frac{n(n+1)(2n+1)}{6} = \frac{100 \times 101 \times 201}{6} = 338350$

Ab values rakhte hain:

\text{Sum of } y_i = 3(338350) - 2(5050)

\text{Sum of } y_i = 1015050 - 10100 = 1004950

4. Final Answer:

\text{Mean of } y = \frac{1004950}{100} = 10049.50

Choose the correct answer:

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