Problem 53 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The variance of first 50 even natural numbers is [2014] (a) 437 (b) \(\frac{437}{4}\) (c) \(\frac{833}{4}\) (d) 833

Step-by-Step Solution

Answer

The variance is 833 (option d).

Step 1: Understanding the Problem

We need to calculate the variance of the first 50 even natural numbers. These numbers are 2, 4, 6, ..., 100.

Step 2: Determine the Sequence

The sequence of the first 50 even natural numbers is an arithmetic sequence where the first term \( a = 2 \) and the common difference \( d = 2 \). The last term \( l = 100 \).

Step 3: Calculate the Mean

The mean \( \bar{x} \) of the numbers is given by \( \bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \), where \( n = 50 \). For an arithmetic sequence, \( \sum = \frac{n}{2}(a+l) \). So, \( \bar{x} = \frac{1}{50} \times \frac{50}{2}(2 + 100) = 51 \).

Step 4: Calculate the Variance Formula

Variance \( \sigma^2 \) is calculated as \( \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \). However, for arithmetic sequences, \( \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2 \).

Step 5: Compute \( \sum_{i=1}^{n} x_i^2 \)

The formula for the sum of squares in an arithmetic sequence is \( \sum_{i=1}^{n} x_i^2 = \frac{n}{6} (2a^2 + (2n-1)ad + n(n-1)d^2) \). For this sequence, this becomes \( \frac{50}{6}(8 + 98 \times 2 + 49 \times 4) = 171700 \).

Step 6: Calculate Variance

Now, plug these into the variance formula. Variance \( \sigma^2 = \frac{1}{50} \times 171700 - 51^2 = 3434 - 2601 = 833 \).

Step 7: Conclusion and Answer

The variance of the first 50 even natural numbers is 833, which matches option (d).

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a series of numbers in which the difference between consecutive terms remains constant. This consistent change between terms is known as the 'common difference', denoted as \(d\). In our problem, we're dealing with the first 50 even natural numbers: 2, 4, 6, ..., 100. The common difference here is 2 since each number increases by 2 from the previous one. Understanding the arithmetic sequence helps simplify calculations because it provides a structured way to describe sequences of numbers, where properties like the mean or sum can be easily calculated.
Key aspects of arithmetic sequences include:

• First term \(a\): This is where the sequence starts. For our even numbers sequence, \(a = 2\).
• Common difference \(d\): This is the fixed interval between terms, which is 2 in this problem.
• Number of terms \(n\): How many terms are there in the sequence. Here, \(n = 50\).

Using these components, we can determine other values, such as the sequence's total and average, which are crucial for solving variance.

Mean of Numbers

The mean or average is a crucial concept that gives us a central value of a series of numbers. For an arithmetic sequence, the mean can be quickly calculated using the formula:\[\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\]where \(\bar{x}\) is the mean, \(n\) is the total number of terms, and \(\sum x_i\) is the sum of the series.
In our example, we calculated \[\bar{x} = \frac{1}{50} \times \frac{50}{2}(2 + 100) = 51\]The formula showcases the power of arithmetic progressions, making it straightforward to compute the mean by merely taking the average of the first and last term in the series. This average, 51, represents the center of our dataset of even numbers.
The mean is essential here because the variance formula, which measures how spread out the numbers are from the mean, requires it.

Sum of Squares

The sum of squares is a sum where each term is squared individually before being added together. It plays a critical role in calculating variance. For an arithmetic sequence, the sum of squares formula is:\[\sum_{i=1}^{n} x_i^2 = \frac{n}{6}(2a^2 + (2n-1)ad + n(n-1)d^2)\]This formula might look complex at first, but it efficiently uses the properties of arithmetic sequences by directly incorporating the first term, common difference, and the number of terms.
In the exercise of finding the variance of the first 50 even natural numbers, this formula simplifies significant computations:\[\sum_{i=1}^{n} x_i^2 = \frac{50}{6}(8 + 98 \times 2 + 49 \times 4) = 171700\]
Once calculated, this sum is instrumental for deriving variance. Understanding the sum of squares gives insights into the distribution's spread by quantifying each deviation from the mean.