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 , … , a 100 is 25 . Then S is [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is $25$ . Then $S$ is [2023]

1 ${99}$

2 $ϕ$

3 $N$

4 ${9}$

Ans.
(3)

$Let a_{1} be a natural number.$

$The 100 consecutive positive numbers are a_{1}, a_{1} + 1, a_{1} + 2, a_{1} + 3, \dots, a_{1} + 99$

$Mean (\bar{x}) = \frac{a_{1} + a_{1} + 1 + a_{1} + 2 + \dots + a_{1} + 99}{100} = a_{1} + \frac{99}{2}$

$Mean deviation about the mean = \frac{\sum_{i = 1}^{100} | x_{_{i}}^{2} - \bar{x} |}{100}$

$= \frac{\frac{99}{2} + \frac{97}{2} + \frac{95}{2} + \dots + \frac{99}{2}}{100} = 25$

$Hence, a_{1} is a natural number.$

Want Us to Email you About Special Offers & Updates?