shyind wrote:

If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

A. A+1

B. A+5

C. A+25

D. 2A

E. 5A

We can label each integer as follows:

First integer = x

Second integer = x + 1

Third integer = x + 2

Fourth integer = x + 3

Fifth integer = x + 4

Thus:

A = 5x + 10

A - 10 = 5x

(A - 10)/5 = x

A/5 - 2 = x

We see that the next 5 integers are x + 5, x + 6, x + 7, x + 8, and x + 9.

Since we have an evenly spaced set, sum = average x quantity, and since x = A/5 - 2, we have:

sum = (first term in the set + last term in the set)/2 x quantity

sum = [(A/5 - 2 + 5 + A/5 - 2 + 9)/2] x 5

sum = [(2A/5 + 10)/2] x 5

sum = (A/5 + 5) x 5

sum = A + 25

Alternate solution:

If we let the first integer be x, then the first five integers will be x, x + 1, x + 2, x + 3, and x + 4, and the next five integers will be x + 5, x + 6, x + 7, x + 8, and x + 9. We can see that each of the next five integers is 5 more than its counterpart in the first five integers. Thus, the sum of the next five integers is 5 x 5 = 25 more than the sum of the first five integers. So, if the sum of the first five integers is A, the sum of the next five integers is A + 25.

Answer: C

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Jeffery Miller

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