a2zinterviews

6.	9, 12, 11, 14, 13, ..., 15
	A.   12	B.   16
	C.   10	D.   17

Solution

16. Alternately we add 3 and subtract 1.

7.	1, 2, 4, 8, 16, 32, 64, ..., 256
	A.   128	B.   148
	C.   154	D.   164

Solution

128. Each Previous number is multiplied by 2.

8.	165, 195, 255, 285, 345, ...
	A.   375	B.   420
	C.   435	D.   390

Solution

435. Each number is multiplied by a prime number i.e., 15x11, 15x13, 15x17 ... So, the next number is 15x29=435.

9.	3, 4, 7, 16, 43, 124, ...
	A.   367	B.   356
	C.   125	D.   1222

Solution

367. Each successive no. is obtained by first multiplying the previous by 3 and then subtracting 5.

10.	3, 7, 6, 5, 9, 3, 12, 1, 15, ...
	A.   18	B.   13
	C.   -1	D.   3

Solution

-1. There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted. The next number is 1 - 2 = -1.