If we take four consecutive natural numbers , what is the relation between the
products of the first and the last , and the product of the middle two ?
Answers
Given : four consecutive natural numbers
To Find : relation between the products of the first and the last , and the product of the middle two
Solution:
four consecutive natural numbers
n , n + 1 , n + 2 , n + 3
products of the first and the last
= n(n + 3)
= n² + 3n
product of the middle two
=(n + 1)(n + 2)
= n² + 3n + 2
= products of the first and the last + 2
product of the middle two = products of the first and the last + 2
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Answer:
Let the four consecutive natural numbers be x,x+1,x+2,x+3.
According to given condition,
x(x+1)(x+2)(x+3)=5040
x(x+3)(x+1)(x+2)=5040
(x
2
+3x)(x
2
+3x+2)=5040
Let x
2
+3x=a
a(a+2)=5040
a
2
+2a=5040
Adding 1 on both sides,
a
2
+2a+1=5040+1
(a+1)
2
=5041
Taking square roots on both sides,
a+1=±71
x
2
+3x−70=0 or x
2
+3x+72=0
The discriminant of the 2
nd
equation is negative, hence no real roots exist for that equation. Hence, we only solve the 1
st
equation.
x(x+10)−7(x+10)=0
(x+10)(x−7)=0
x+10=0 or x−7=0
x=−10 or x=7
−10 is not a natural number
∴x=−10 is not applicable
The numbers are 7,7+1=8,7+2=9,7+3=10
∴ The numbers are 7,8,9,10.