Question

Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of $100$ consecutive positive integers $a_1, a_2, a_3, \ldots , a_{100}$ is $25$ Then $S$ is

• A. $\{99\}$
• B. $\phi$
• C. $N$
• D. $\{9\}$

Answer 1

The Correct Option is C

We need to determine the set \(S\) which represents the values of \(a_1\) for which the mean deviation about the mean of 100 consecutive positive integers is 25.

To solve this, let's start by defining the sequence of consecutive integers:

• \(a_1, a_2, \ldots, a_{100}\) are 100 consecutive integers.
• If \(a_1\) is the first integer, the integers are \(a_1, a_1 + 1, a_1 + 2, \ldots, a_1 + 99\).

The mean \(M\) of these numbers can be calculated as follows:

\(M = \frac{1}{100} \sum_{i=1}^{100} (a_1 + (i-1)) = \frac{1}{100} \left(100a_1 + \frac{99 \times 100}{2}\right)\)

Simplifying, we get:

\(M = a_1 + \frac{99}{2} = a_1 + 49.5\)

Now, the mean deviation about the mean \(M\) is given by:

\(\frac{1}{100} \sum_{i=1}^{100} |a_i - M| = 25\)

Expanding this for our sequence:

\(\frac{1}{100} \sum_{i=0}^{99} |a_1 + i - (a_1 + 49.5)| = 25\)

which simplifies to:

\(\frac{1}{100} \sum_{i=0}^{99} |i - 49.5| = 25\)