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Seven consecutive odd integers have a sum of 441. Find the highest integer.

A. 63
B. 65
C. 67
D. 69
E. 71​

Sagot :

Answer:

To determine the highest integer among seven consecutive odd integers whose sum is 441, we first define the sequence. Let the middle integer of the sequence be ( x ). Since we are dealing with seven consecutive odd integers, the sequence can be written as:

[x-6, x-4, x-2, x, x+2, x+4, x+6]

Next, we sum these integers:

[(x-6) + (x-4) + (x-2) + x + (x+2) + (x+4) + (x+6)]

Combining the terms, we get:

[(x-6) + (x-4) + (x-2) + x + (x+2) + (x+4) + (x+6) = 7x]

Given that the sum is 441, we set up the equation:

[7x = 441]

Solving for \( x \):

[x = \frac{441}{7} = 63]

Now, since \( x \) is the middle integer in the sequence, the highest integer is:

[x + 6 = 63 + 6 = 69]

Therefore, the highest integer is D. 69

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