MCQMediumJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

Let SS be the set of all values of a1a_1 for which the mean deviation about the mean of 100100 consecutive positive integers a1,a2,a3,,a100a_1, a_2, a_3, \dots, a_{100} is 2525. Then SS is:

  • A

    \emptyset

  • B

    {99}\{99\}

  • C

    N\mathbb{N}

  • D

    {}\{\}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The numbers are 100100 consecutive positive integers starting from a1a_1, so they are a1,a1+1,a1+2,,a1+99a_1, a_1+1, a_1+2, \dots, a_1+99.

Find: The set of all values of a1a_1 for which the mean deviation about the mean is 2525.

The solution states that a1a_1 can be any natural number.

First compute the mean:

xˉ=a1+(a1+1)+(a1+2)++(a1+99)100\bar{x} = \frac{a_1 + (a_1+1) + (a_1+2) + \dots + (a_1+99)}{100} =100a1+(1+2++99)100= \frac{100a_1 + (1+2+\dots+99)}{100} =a1+992= a_1 + \frac{99}{2}

Now the mean deviation about the mean is

1100i=1100xixˉ\frac{1}{100}\sum_{i=1}^{100} |x_i-\bar{x}|

For these equally spaced consecutive integers, the deviations from the mean are symmetric, so the expression becomes

2100(992+972+952++12)\frac{2}{100}\left(\frac{99}{2} + \frac{97}{2} + \frac{95}{2} + \dots + \frac{1}{2}\right) =1100(99+97+95++1)= \frac{1}{100}(99+97+95+\dots+1)

The sum of the first 5050 odd numbers is

1+3+5++99=502=25001+3+5+\dots+99 = 50^2 = 2500

Hence

Mean deviation=2500100=25\text{Mean deviation} = \frac{2500}{100} = 25

This value is independent of a1a_1. Therefore, the condition is satisfied for every natural number a1a_1. The correct option is C.

Translation Invariance Trick

Given: The numbers are consecutive integers shifted by a1a_1.

Find: Whether the mean deviation depends on a1a_1.

Adding the same constant to every observation shifts the mean by the same amount, but leaves all deviations from the mean unchanged. Therefore, the mean deviation about the mean for

a1,a1+1,,a1+99a_1, a_1+1, \dots, a_1+99

is the same as that for

1,2,3,,1001,2,3,\dots,100

For 100100 consecutive integers, the data are symmetric about the mean, and the mean deviation evaluates to 2525. Hence it is true for every natural number a1a_1, so S=NS=\mathbb{N}. The correct option is C.

Common mistakes

  • Assuming the mean deviation depends on a1a_1. This is wrong because shifting every term by the same constant shifts the mean equally and does not change the absolute deviations. Instead, focus on the spacing of the numbers, not their starting point.

  • Using the mean as a1+50a_1+50 or a1+49a_1+49. This is wrong because the mean of 100100 consecutive integers from a1a_1 to a1+99a_1+99 is the average of the first and last terms, namely a1+992a_1+\frac{99}{2}. Always use midpoint of the first and last term.

  • Summing the deviations incorrectly by ignoring symmetry. This is wrong because the deviations occur in equal pairs on both sides of the mean. Instead, write one side of the deviations and double it.

Practice more Sum of Series questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions