Show solving all the integer groups one by one.
2007-03-16 17:53:37 · 8 answers · asked by Kritika Srivastav 2 in Science & Mathematics ➔ Mathematics
Sorry I had mistaken the spelling that is consecutive.
2007-03-16 17:54:49 · update #1
The general rule is the sum of three consecutive numbers is eual to 3 × middle numbet in those 3 numbers.
Similarly sum of five consecutive numbers is eual to 5 × middle numbet in those 5 numbers.
and so on.
Here the word 'between' indicates that we have to take numbers from 2 to 9
forming the groups
2+3+4 = 9 = 3 × 3 (middle no.)
3+4+5 = 12 = 3 × 4 (middle no.)
4+5+6 = 15 = 3 × 5 (middle no.)
5+6+7 = 18 = 3 × 6 (middle no.)
6+7+8 = 21 = 3 × 7 (middle no.)
7+8+9 = 24 = 3 × 8 (middle no.)
2007-03-17 05:11:08 · answer #1 · answered by Pranil 7 · 0⤊ 0⤋
If the integers are between 1 to 10 then they can only be 2, 3, 4, 5, 6, 7, 8, and 9.
2+3+4=9
3+4+5=12
4+5+6=15
5+6+7=18
6+7+8=21
7+8+9=24
2007-03-16 18:36:43 · answer #2 · answered by Felicia P 2 · 0⤊ 0⤋
1+2+3 = 6
2+3+4 = 9
3+4+5 = 12
4+5+6 = 15
5+6+7 = 18
6+7+8 = 21
7+8+9 =24
8+9+10 = 27
the pattern is the (middle number*3)
To show this
let first integer be n
therefore the middle would n+1 and the third would be n+2
s = n+(n+1)+(n+2)
s = 3n+3
s = 3(n+1)
s = 3(n+1), since n+1 is the middle number it's 3*middle number
or since n is the first number you can say it's 3 times the first number plus s = 3n+3
or you can find it relative to the last number
s=3n+3
add 3 and subtract 3
s=3n+6 - 3
s =3(n+2) - 3
so given n+o+p the sum is
3n+3 where n is the first number or
3o where o is the middle number or
3p-3 where p is the last number.
2007-03-16 18:00:12 · answer #3 · answered by radne0 5 · 1⤊ 0⤋
1+2+3 = 6
2+3+4 = 9
3+4+5 = 12
4+5+6 = 15
5+6+7 = 18
6+7+8 = 21
7+8+9 = 24
8+9+10 = 27
Looks like each number is three more than the last, and after reading the answer above mine, it looks as if he is right. The middle number multiplied by three is the number you are looking for.
2007-03-16 17:58:21 · answer #4 · answered by HallamFoe 4 · 1⤊ 0⤋
if a= first number
a+1=second number
a+2=third number
if a=1 we have 1+2+3=6
if we have a+2=10=>a=8 so 8+9+10=27
6<=sum<=27
2007-03-16 18:09:00 · answer #5 · answered by djin 2 · 0⤊ 0⤋
Sum of three consecutive integers are greater than 6 and lessthan
27.
2007-03-16 17:59:59 · answer #6 · answered by Bharath P 1 · 0⤊ 0⤋
The minimum sum can be 6 and the maximum can be 27 .
2007-03-16 18:01:03 · answer #7 · answered by ritesh s 2 · 0⤊ 0⤋
I think it's the middle number multiplyed by three. Yeah, this works with all numbers.
ex- 3 4 5=12(4x3)
2007-03-16 17:57:45 · answer #8 · answered by Anonymous · 2⤊ 0⤋