Question

Product of 4 consecutive natural numbers is 840 find no.

Answer 1

let the numbers are n,n+1,n+2,n+3

given-product is 840

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

n^4+6n^3+11n^2+6n-840=0

solving the equation we get n=4

therefore the numbers are 4,5,6,7

Answer 2

Answer:

Consecutive natural nos. are 4 , 5 , 6 , 7

Step-by-step explanation:

Given: Product of 4 consecutive natural no. = 840

To find: Four Nos.

Let the four consecutive natural nos are x , x + 1 , x + 2  &  x + 3

According to the Question,

x × ( x + 1 ) × ( x + 2 ) × ( x + 3 ) = 840

( x² + x ) ( x² + 5x + 6 ) = 840

$x^4+6x^3+11x^2+6x=840$

$x^4+6x^3+11x^2+6x-840=0$

let, $f(x)=x^4+6x^3+11x^2+6x-840$

we find value of x by hit & trial method

for x = 1

$f(1)=1^4+6\times1^3+11\times1^2+6\times1-840=24-840=-816$

for x = 2

$f(2)=2^4+6\times2^3+11\times2^2+6\times2-840=120-840=-720$

for x = 3

$f(3)=3^4+6\times3^3+11\times3^2+6\times3-840=360-840=-480$

for x = 4

$f(4)=4^4+6\times4^3+11\times4^2+6\times4-840=840-840=0$

⇒ x = 4  is first no.

Nos. are consecutive,

⇒ Required nos = 4 , 5 , 6 , 7

Therefore, Consecutive natural nos. are 4 , 5 , 6 , 7