The remainder obtained when 1! + 2! + ... + 95! is divided by 15 is
14
3
1
0
B.
3
Since, 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120
Since, all terms from 5! onwards are divisible by 15 and 1! + 2! + 3! + 4! = 33
The required remainder after dividing by 15 will be 3
If a, b and c are in arithmetic progression, then the roots of the equation ax - 2bx + c = 0 are
Let the coefficients of powers of x in the 2nd, 3rd and 4th terms in the expansion of (1 + x)n, where n is a positive integer, be in arithmetic progression. Then, the sum of the coefficients of odd powers of x in the expansion is
32
64
128
256
Six numbers are in AP such that their sum is 3. The first term is 4 times the third term. Then, the fifth term is
- 15
- 3
9
- 4