Let the six numbers a 1 , a 2 , a 3 , a 4 , a 5 , a 6 be in A.P. and a 1 + a 3 = 10 . If the mean of these six numbers is 19 2 and their variance is 2 , then 8 2 is equal to [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 the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to [2023]

1 105

2 210

3 200

4 220

Ans.
(2)

$Given a_{1}, a_{2}, a_{3}, \dots, a_{6} are in A.P.$

$Sum of these terms in A.P. = \frac{6}{2} [2 a + 5 d] = \frac{19}{2} \times 6$

$\Rightarrow 2 a + 5 d = 19 ...(i)$

$Since a_{1} + a_{3} = 10 \Rightarrow a + a + 2 d = 10$

$\Rightarrow 2 a + 2 d = 10 ...(ii)$

$Solving (i) and (ii), we get a = 2, d = 3$

$So, 2, 5, 8, 11, 14, 17 are the terms.$

$Variance, σ^{2} = \frac{2^{2} + 5^{2} + 8^{2} + 11^{2} + 14^{2} + 17^{2}}{6} - {(\frac{19}{2})}^{2}$

$= \frac{699}{6} - \frac{361}{4} = \frac{466 - 361}{4} = \frac{105}{4}$ $\Rightarrow 8 σ^{2} = \frac{8 \times 105}{4} = 210$

Want Us to Email you About Special Offers & Updates?