the sum of four consecutive odd intergers is 64. what are the integers thanks
your 4 odd integers are:
2n+1, 2n+3, 2n+5, 2n+7
they are such that
(2n+1)+(2n+3)+(2n+5)+(2n+7)=64
8n+16=64
8n=64-16=48
n=48/8=6
then your numbers are
2*6+1, 2*6+3, 2*6+5, 2*6+7
13, 15, 17 and 19
x+x+2+x+4+x+6=64
so
4x+12=64
subtract 12 from both sides
4x=52
divide both sides by 4
x=13
so the four consecutive odd integers are:
13, 15, 17, 19
1st integer—x:
x + x + 2 + x + 4 + x + 6 = 64
4x = 52
x = 13
2nd integer:
= 13 + 2
= 15
3rd integer:
= 13 + 4
= 17
4th integer:
= 13 + 6
= 19
Answer: 13, 15, 17 & 19 are the integers. Their sum is 64.
Let the odd numbers be
2k-3, 2k-1, 2k+1 and 2k+3
So sum= 8k= 64
k=8
Numbers are
13,15,17 and 19
the equation is set up like this:
x+(x+2)+(x+4)+(x+6) =64
integers are 13,15,17,19
Comments
your 4 odd integers are:
2n+1, 2n+3, 2n+5, 2n+7
they are such that
(2n+1)+(2n+3)+(2n+5)+(2n+7)=64
8n+16=64
8n=64-16=48
n=48/8=6
then your numbers are
2*6+1, 2*6+3, 2*6+5, 2*6+7
13, 15, 17 and 19
x+x+2+x+4+x+6=64
so
4x+12=64
subtract 12 from both sides
4x=52
divide both sides by 4
x=13
so the four consecutive odd integers are:
13, 15, 17, 19
1st integer—x:
x + x + 2 + x + 4 + x + 6 = 64
4x = 52
x = 13
2nd integer:
= 13 + 2
= 15
3rd integer:
= 13 + 4
= 17
4th integer:
= 13 + 6
= 19
Answer: 13, 15, 17 & 19 are the integers. Their sum is 64.
Let the odd numbers be
2k-3, 2k-1, 2k+1 and 2k+3
So sum= 8k= 64
k=8
Numbers are
13,15,17 and 19
the equation is set up like this:
x+(x+2)+(x+4)+(x+6) =64
4x+12=64
4x=52
x=13
integers are 13,15,17,19