Question

There are 6 consecutive odd natural numbers in increasing order. The difference between the average of the squares of the first 4 numbers and the last four numbers is 64. If the sum of the squares of the first and the last element (i. e , odd numbers) is 178, then the average of all the six numbers is :

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The correct option is B 8

Let the numbers be $(a-5),(a-3),(a-1),(a+1),(a+3).(a+5)$
then their average
$\frac{(a-5)+(a-3)+(a-1)+(a+1)+(a+3)+(a+5)}{6}=a$
Again the value of 'a' can be found by using the last statement
i.e., $(a-5{)}^2+(a+5{)}^2=178\Rightarrow a^2=64\Rightarrow a=8$