Four times the first of three consecutive odd integers is 6 more than twice the third The third integer is

11: let no. be a-2 , a , a+2 be three consecutive odd integers. then 4(a-2) = 6 + 2(a+2)
solving we get a=9;
third no. : a+2 so 9+2=11;
7 years agoHelpfull: Yes(4) No(1)
d)11
4(7)=(6+11*2)
Here_the three consecutive ood integers are 7,9,11.
7 years agoHelpfull: Yes(2) No(1)
Ans is 11
4(2n+1)=2(2n+5)+6 solving this we get n=3.
So third integer is 11
6 years agoHelpfull: Yes(1) No(0)
let 3 consecutive odd number be:
(2n+1), (2n+3), (2n+5)

according to given condition:
3(2n+1)=2(2n+5) + 6
therefore n=3

so 3 numbers are 7,9,11
3 years agoHelpfull: Yes(1) No(0)
Take the odd number is n,
next odd number is n+2,
next odd number is n+4
4n=2(n+4)+6
4n=2n+14
n=7
n+4=11
7 years agoHelpfull: Yes(0) No(0)
11

c=a+4
according to que
4a = 2c+6
after solving we get c = 11
7 years agoHelpfull: Yes(0) No(0)
11 is the answer
5 years agoHelpfull: Yes(0) No(0)
Let us assume the three odd integers
a,a+2,a+4
Given four times the first of three consecutive odd integers so…
4a= 6+2(a+4)
4a = 6+2a+8
4a-2a = 14
2a = 14
a = 7
Hence
a+4= 7+4=11
11 is the answer
2 years agoHelpfull: Yes(0) No(0)
a, a+2 ,a+4
4a = 6+2 (a + 4)
4a = 6+ 2a + 8
4a - 2a = 14
2a = 14
a = 7
hence
a + 4 = 7 + 4 = 11
hence proved
1 year agoHelpfull: Yes(0) No(0)

Four times the first of three consecutive odd integers is 6 more than twice the third. The third integer is
a)7
b)9
c)13
d)11
e)15