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(B) Let the $6$ consecutive odd numbers be $(a-5), (a-3), (a-1), (a+1), (a+3), (a+5)$.The average of all $6$ numbers is $\frac{(a-5)+(a-3)+(a-1)+(a+1)

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the $6$ consecutive odd numbers be $(a-5), (a-3), (a-1), (a+1), (a+3), (a+5)$.The average of all $6$ numbers is $\frac{(a-5)+(a-3)+(a-1)+(a+1)+(a+3)+(a+5)}{6} = a$.Given that the sum of the squares of the first and last element is $178$:$(a-5)^2 + (a+5)^2 = 178$$(a^2 - 10a + 25) + (a^2 + 10a + 25) = 178$$2a^2 + 50 = 178$$2a^2 = 128$$a^2 = 64$$a = 8$ (since the numbers are positive).Verification with the first condition:Average of squares of first $4$ numbers: $\frac{(a-5)^2 + (a-3)^2 + (a-1)^2 + (a+1)^2}{4} = \frac{(3)^2 + (5)^2 + (7)^2 + (9)^2}{4} = \frac{9 + 25 + 49 + 81}{4} = \frac{164}{4} = 41$.Average of squares of last $4$ numbers: $\frac{(a-1)^2 + (a+1)^2 + (a+3)^2 + (a+5)^2}{4} = \frac{(7)^2 + (9)^2 + (11)^2 + (13)^2}{4} = \frac{49 + 81 + 121 + 169}{4} = \frac{420}{4} = 105$.Difference: $105 - 41 = 64$. This matches the given condition.Therefore,the average of all $6$ numbers is $a = 8$.

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 $4$ numbers is $64$. If the sum of the squares of the first and the last element is $178$,then the average of all the $6$ numbers is:

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