Problem 65 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the mean deviation of the numbers \(1,1+\mathrm{d}\), \(1+2 \mathrm{~d}, \ldots .1+100 d\) from their mean is 255 , then \(d\) is equal to: [2009] (a) \(20.0\) (b) \(10.1\) (c) \(20.2\) (d) \(10.0\)

Step-by-Step Solution

Answer

The value of \(d\) is 10.0.

Step 1: Understand the Sequence

The sequence given is an arithmetic sequence with the first term as 1 and the common difference as \(d\). The terms are \(1, 1+d, 1+2d, \ldots, 1+100d\).

Step 2: Compute the Mean

The mean of an arithmetic sequence can be found using the formula for the average of the first and last term since the terms are evenly spaced. Here, the mean \(\bar{x}\) is \[\bar{x} = \frac{(1) + (1 + 100d)}{2} = 1 + 50d.\]

Step 3: Calculate Deviations from the Mean

For each term \(1 + kd\), the deviation from the mean is \((1 + kd) - (1 + 50d) = (k - 50)d\). The terms vary from \(1 + 0d\) to \(1 + 100d\).

Step 4: Compute the Mean Deviation

The mean deviation is the average of the absolute deviations from the mean. This is given by \[\frac{1}{101} \sum_{k=0}^{100} |(k-50)d|.\] Given that this mean deviation is 255, we set up the equation: \(\frac{1}{101} \sum_{k=0}^{100} |(k-50)d| = 255\).

Step 5: Simplify and Solve for \(d\)

Notice that from \(k=0\) to \(k=50\), \(|k-50| = 50-k\) and from \(k=51\) to \(k=100\), \(|k-50| = k-50\). Calculate each range separately: \[ \sum_{k=0}^{50} (50-k) = \sum_{k=0}^{50} k = \frac{50 \times 51}{2}, \] \[ \sum_{k=51}^{100} (k-50) = \sum_{k=1}^{50} k = \frac{50 \times 51}{2}. \] Add these sums to find \(255 \times 101 = 50d \times 51\). Solve for \(d\): \(d = \frac{255 \times 2}{51} = 10.\)

Step 6: Verification

Verify by plugging \(d = 10\) back into the context of the problem to ensure the mean deviation results in 255 as expected.

Key Concepts

Arithmetic Sequence

An arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers in which each term after the first is formed by adding a constant, called the "common difference" (denoted by \( d \)), to the previous term. This constant difference is what generates the sequence's linear nature.

• For example, in the sequence given by the problem, the first term is 1, and each subsequent term is found by adding \( d \). This creates the sequence: \( 1, 1+d, 1+2d, \ldots, 1+100d \).
• The common difference \( d \) helps determine the spread and positioning of the terms within the sequence.
• The last term of this sequence can be expressed as \( 1+100d \).

Understanding arithmetic sequences aids in calculating various statistics, such as the mean, effectively.

Deviation from Mean

Deviation from mean is the difference between each term in a sequence and the average (mean) of the sequence. In mathematical contexts, especially involving series and arithmetic sequences, understanding deviations aids in determining the spread or variability of a sequence of numbers.

• To find each deviation, subtract the mean of the sequence from each term.
• In our arithmetic sequence, if the mean (\( \bar{x} \)) is \( 1 + 50d \), then each term \( 1+kd \) has a deviation of \( (k-50)d \).
• This approach shows the distance of each term from the center, indicating whether each term is above or below the average.

The calculation of deviation gives insights into how concentrated or dispersed the terms are around the mean.

Problem-Solving in Mathematics

Problem-solving in mathematics involves breaking down complex problems into manageable steps. This systematic approach helps in understanding and solving mathematical questions efficiently.

• Step 1: Understand the problem by identifying the sequence and terms.
• Step 2: Compute the mean or center of your sequence to set a foundation.
• Step 3: Evaluate deviations to measure variability.
• Step 4: Solve any equations or expressions set up from the problem definitions.
• Step 5: Verify the solution by plugging it back into the context or checking against conditions given in the problem.

Applying these steps helps students to systematically approach mathematical questions, such as the mean deviation problem, ensuring their strategy is thorough and logical.

JEE Main Mathematics

JEE Main is an entrance examination in India for students seeking admission in various undergraduate engineering programs. The JEE Main Mathematics section tests concepts like arithmetic sequences and deviations.

• These types of problems frequently appear, requiring students to have deep understanding and quick problem-solving abilities.
• The mean deviation problem serves as an excellent example of typical questions within the syllabus that test analytical and computational skills.
• Understanding fundamental concepts and practicing a variety of problems is vital for students preparing for such competitive exams.

With practice, students can effectively tackle mathematical problems and excel in exams like JEE Main, which are crucial for their academic and professional futures.