The product of four constitutive natural number is 840 find the number
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12 , 10 , 8 is ur ans
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Here is your answer :-
Let four consecutive number are n , n+1 , n+2 and n+3
ATQ ,
n(n+1)(n+2)(n+3) = 840
n(n+1)(n² + 5n + 6 ) = 840
( n² + n )( n²+5n+6 ) = 840
n⁴ +5n³ + 6n² + n³ + 5n² + 6n = 840
n⁴ + 6n³ + 11n² + 6n = 840
(n+1) = 840
n = 839
Let four consecutive number are n , n+1 , n+2 and n+3
ATQ ,
n(n+1)(n+2)(n+3) = 840
n(n+1)(n² + 5n + 6 ) = 840
( n² + n )( n²+5n+6 ) = 840
n⁴ +5n³ + 6n² + n³ + 5n² + 6n = 840
n⁴ + 6n³ + 11n² + 6n = 840
(n+1) = 840
n = 839
chirag1212:
x x+1 x+2 x+3 =840 x^2+3x x^2+3x+2 =840 put x^2+3x =p
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