If l, m, n are in arithmetic progression, then the straight line b + my + n = 0 will pass through the point
(- 1, 2)
(1, - 2)
(1, 2)
(2, 1)
Let n = 1! + 4! + 7! + . . . + 400!. Then ten's digit of n is
1
6
2
7
B.
6
Given, n = 1! + 4! + 7! + ... + 400!
1! = 1, 4! = 24, 7! = 5040, 10! = 3628800
and further the terms has last two digits = 00
So, n = 1! + 4! + 7! + ... + 400! = . . . 65
Here ten digit is 6.
Hence, 6 is the answer
Let a = for n = 1, 2, 3 . . . then the greatest value of n for which an is the greatest is
11
20
10
8