WILL GIVE BRAINLIEST ANSWER IF ANSWER WITH EXPLANATION Given that the three factorials n!, (n+1)! and (n+19) · n! form an arithmetic sequence where n is a positive integer, what is the product of the digits of n?

The product of the digits of n is 8

Step-by-step explanation:

An arithmetic series is of the form

The nth term of an arithmetic series = aₙ + (n - 1)d

Therefore, for the series, we have;

n!, (n+1)!, (n+19)n!

Therefore;

n! + (n - 1)·d = (n+1)! = (n + 1)·n! = n·n! + n!

n·n! = (n - 1)·d

n! + (n + 1 - 1)·d =  (n+19)n!

n! + n·d =  n·n!+19·n! = (n - 1)·d +19·n! = n·d - d + 19·n!

n! - 19·n! = n·d - n·d - d

-18·n! = -d

Therefore, d = 18·n!

n! + d = (n + 1)!

n! + 18·n! = (n + 1)·n!

n!·(18 + 1) = (n + 1)·n!

18 + 1 = n + 1

∴ n = 18

Hence the product of the digits of n = 1 × 8 = 8

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation: