There are 6 consecutive odd numbers in increasing order. difference between average of squares of fi

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There are $6$ consecutive odd 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

$10$

Solution

let the number be $(a-5),(a-3),(a-1),(a+1)(a+3)(a+5)$

Then the average of all six consecutive odd no.

$=\frac{(a-5)+(a-3)+(a-1)+(a+1)+(a+3)+(a+5)}{6}=a$

The value of "a' can be found by using the last statement $(a-5)^{2}+(a+5)^{2}=178 \Rightarrow a^{2}=64 a=8$

Average of all six number is $8.$

Standard 13

Quantitative Aptitude

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