I am thinking of four consecutive integers. The sum of the second and fourth equals the third increased by 17.
Write that into an equation.
2006-11-06 05:43:44 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Consecutive integers are n, n + 1, n + 2, n + 3
The sum of the second and fourth equals the third increased by 17.
(n + 1) + (n + 3) = (n + 2) + 17
2006-11-06 05:51:50 · answer #1 · answered by Anonymous · 1⤊ 1⤋
integers are :n, n+1, n+2, n+3
n++1+n+3=n+2+17
2n+4=n+19
n=15
the integers are 15, 16, 17, 18
16+18=34
17=17=34
2006-11-06 14:23:53 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋
suppose the first one is n then teh numbers are:
n, (n+1), (n+2), (n+3)
sum of second and forth one is (n+1)+(n+3) and equal (n+2)+17
(n+1)+(n+3)=(n+2)+17 then
2n+4=n+19 and
n=15 so the numbers are
15, 16, 17, 18
and 16+18=17+19
2006-11-06 14:27:52 · answer #3 · answered by Totok 2 · 0⤊ 0⤋
the integers be n,n+1,n+2,n+3
(n+1)+(n+3)=(n+2)+17
2006-11-06 13:47:49 · answer #4 · answered by raj 7 · 1⤊ 0⤋
n,n+1,n+2, n+3, four consecutive integers
(n+1)+(n+3)=(n+2)+17
2n+4=n+19
n=15,
15,16,17,18
check
16+18=34
17+17=34
2006-11-06 13:58:07 · answer #5 · answered by chanljkk 7 · 0⤊ 0⤋
d=c+1=b+2=a+3
and you want
b+d=c+17
so solve for b and d
c+1=b+2
c-1=b
c+1=d
so you substitute:
c-1+c+1=c+17
2c-c=17
c=17
and then a=15,b=16,c=17,d=18
2006-11-06 13:50:22 · answer #6 · answered by nemahknatut88 2 · 0⤊ 0⤋