Answer:
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
