Question

The product of four consecutive natural numbers
is 84o. find the numbers.​

Answer 1

Answer:

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Step-by-step explanation:

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.

Answer 2

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❏Given:

The Product Of 4 Consecutive Natural Numbers=840

❏To Find:

Numbers

❏Formula:

Let the consecutive numbers be n, (n+1) , (n + 2 ) (n+ 3).

Given that their product = 840

then n(n+1)(n + 2 )(n+ 3) = 840.

this can also written as n(n+3)⋅(n+1)(n+2) = 840

Add and subtarct 1 to the first number of the product.

((n^2+3n+1)−1)((n2+3n+1)+1) = 840

(n^2+3n+1)^2−1 = 840

(n^2+3n+1)^2 = 841

(n^2+3n+1)^2 = 29 ^2

(n^2+3n+1) = 29 implies n^2+3n+1–29=0 = n^2 + 3n - 28

(n + 7) (n - 4) = 0

if n = 4 we have 4, 5, 6, 7 n = -7 is not possible -7 does not belongs to the natural number.

HOPE IT HELPS U! :)

Have a Nice Day Ahead!