Qn: The average of 7 positive consecutive integers is 6. find average of square of these integers
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Answer :
Let us consider that the 7 consecutive integers are n, (n + 1), (n + 2), (n + 3), (n + 4), (n + 5) and (n + 6), where n ∈ |N
The sum of the 7 consecutive integers
= n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5) + (n + 6)
= n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6
= 7n + 21
By the given condition,
(7n + 21)/7 = 6
⇒ n + 3 = 6
⇒ n = 6 - 3
⇒ n = 3
Now, the sum of the squares of these integers
= n² + (n + 1)² + (n + 2)² + (n + 3)² + (n + 4)² + (n + 5)² + (n + 6)²
= 3² + (3 + 1)² + (3 + 2)² + (3 + 3)² + (3 + 4)² + (3 + 5)² + (3 + 6)²
= 3³ + 4² + 5² + 6² + 7² + 8² + 9²
= 9 + 16 + 25 + 36 + 49 + 64 + 81
= 280
∴ the average of squares of these integers
= 280/7
= 40
#MarkAsBrainliest
Let us consider that the 7 consecutive integers are n, (n + 1), (n + 2), (n + 3), (n + 4), (n + 5) and (n + 6), where n ∈ |N
The sum of the 7 consecutive integers
= n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5) + (n + 6)
= n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6
= 7n + 21
By the given condition,
(7n + 21)/7 = 6
⇒ n + 3 = 6
⇒ n = 6 - 3
⇒ n = 3
Now, the sum of the squares of these integers
= n² + (n + 1)² + (n + 2)² + (n + 3)² + (n + 4)² + (n + 5)² + (n + 6)²
= 3² + (3 + 1)² + (3 + 2)² + (3 + 3)² + (3 + 4)² + (3 + 5)² + (3 + 6)²
= 3³ + 4² + 5² + 6² + 7² + 8² + 9²
= 9 + 16 + 25 + 36 + 49 + 64 + 81
= 280
∴ the average of squares of these integers
= 280/7
= 40
#MarkAsBrainliest
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Answer:
282/7
Step-by-step explanation:
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