Find five consecutive odd integers such that the sum of the first and the fifth is one less than three times the fourth

Question

 Please tell me how to solve this problem and all the steps required with it.

Answer 1

Hi Michael.

 

First of all, if you need ODD numbers, you will need 2n+1  (the 2n makes it even, then adding one guarantees it is odd.)

 

First odd integer is 2n+1

2nd odd integer is 2n+3

3rd odd integer is 2n+5

4th odd integer is 2n+7

5th odd integer is 2n+9

 

So if the sum of the first and fifth  [(2n+1) + (2n + 9)]

is one less than 3 times the fourth  [3(2n+7)-1)

 

Set these two expressions equal and find n.  Remember, n is NOT the final answer, 2n+1 is your first odd integer.

 

2n+1+2n+9=6n+21-1

4n+10=6n+20

 

-2n=10

 

n=-5

 

so the first odd integer is 2(-5)+1 = -10+1=-9

 

first = -9

second = -7

third = -5

fourth = -3

fifth = -1

 

Now lets check:  1st + 5th = -10

3 * fourth -1` = 3(-3) - 1 = -9 -1 = -10  YAY!!!!  It works.

Answer 2

n, n+2, n+4, n+6, n+8

 

n = n + 8 = -1 +3(n + 6)

 

Can you solve for n and answer?