Three Consecutive integers are as such they
are taken in increasing onder and multiplied
by 2, 3, ond 4 respectively, they add up to 56. find
these numbers.
Answers
First, let's name the three consecutive integers.
Let's call the first integer:
n
Then the next two integers will be
(
n
+
1
)
and
(
n
+
2
)
If we then multiply them as described in the problem and sum these products to 56 we can write an equation as:
2
n
+
3
(
n
+
1
)
+
4
(
n
+
2
)
=
56
We can now solve this equation for
n
:
2
n
+
(
3
×
n
)
+
(
3
×
1
)
+
(
4
×
n
)
+
(
4
×
2
)
=
56
2
n
+
3
n
+
3
+
4
n
+
8
=
56
2
n
+
3
n
+
4
n
+
3
+
8
=
56
(
2
+
3
+
4
)
n
+
(
3
+
8
)
=
56
9
n
+
11
=
56
9
n
+
11
−
11
=
56
−
11
9
n
+
0
=
45
9
n
=
45
9
n
9
=
45
9
9
n
9
=
5
n
=
5
Therefore:
n
+
1
=
5
+
1
=
6
n
+
2
=
5
+
2
=
7
The three consecutive integers are: 5, 6, 7
Answer:Three consecutive integers are 7,8 and 9
Solution:
Let the numbers are x, x+1 and x+2
(Since these are consecutive integers)
they are taken in increasing order and multiplied by 2,3 and 4 respectively
So x becomes 2x
(x+1) becomes 3(x+1)
(x+2) becomes 4(x+2)
they add up to 74,so ATQ
2x+3(x+1)+4(x+2)=74
2x+3x+3+4x+8=74
9x+11=74
9x=74−11
9x=63
x=
9
63
x=7
So,
Integers are 7,8,9