SOLUTION: Find five consecutive integers such that: “The sum of the first and 4 times the third is equal to 56 less than 3 times the sum of the second, fourth, and fifth.”

Question 1203707: Find five consecutive integers such that:
“The sum of the first and 4 times the third is equal to 56 less than 3 times the sum of the second, fourth, and fifth.”

Found 3 solutions by josgarithmetic, MathLover1, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Mostly you write the worded description as symbolism literally.

Not the only way to assign, but:
n-2
n-1
n
n+1
n+2

Solve this....

---------------the integer in the middle

**************************this is wrong******************************

-----------NO. This does not check.
MISTAKE SOMEWHERE....
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

The formula for a sequence of consecutive integers is , , ,, ,..., .
The sum of the first and times the third is equal to less than times the sum of the second, fourth, and fifth.

five consecutive integers are:, , , ,
check:

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Find five consecutive integers such that:
“The sum of the first and 4 times the third is equal to 56 less than 3 times
the sum of the second, fourth, and fifth.”
~~~~~~~~~~~~~~~~

Let the numbers n, (n+1), (n+2), (n+3) and (n+4) be five consecutive integer numbers. Write equation as you read the problem n + 4*(n+2) = 3*((n+1) + (n+3) + (n+4)) - 56. Simplify step by step n + 4n + 8 = 3*(3n+8) - 56 5n + 8 = 9n + 24 - 56 8 - 24 + 56 = 9n - 5n 40 = 4n n = 40/4 = 10. ANSWER. The numbers are 10, 11, 12, 13, 14.

Solved.

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