NUMBER SEQUENCES for practice for BITSAT

1. answer 21 since all others r multiples of 2
2. it shuld have been 50
since 5-1 = 4 = 2² , 14-5 = 9 = 3², 30 - 14 = 16=4², 55-30 = 25=5² , 91 - 55 = 36 = 6² hence 50 can no where be included here
5. 64 is odd
since the differences increase like this 3 , 5 , 7 , 11 , 13 , ......
but 64 - 50 = 14 which can't be included
6. similar way as abv odd is 102
since 102 - 84 = 18 , but differences increase like this 9 , 11 , 13 ,15, 17 ,.....
8. 18 is odd since
5-4 = 1
7-5 = 2.....................and hence diffrences increaeses like 1 , 2 , 3 , 4 .......... so 18 is fit in the series instead 18 it shuld have been 19

Is something left??...aha 7) 324→ sum of digits is not 10

And 4)→b

Product of extreme terms is not equal to to middle term.

Done!![4]

Reasons according to the book...

(1-8)

1)each except 21 is even number

2)The pattern is 1² , 1²+2² , 1²+2²+3² ,....(50 is not of this pattern)

3)In each except 427 , the midlle digit is a sum of the other two

4)In each except 383 , the product of the extreme terms=midlle term

5)The pattern is x²+1 ...64 out of pattern

6)The pattern is x²+3.. 102 out of pattern

7)Sum of digits in each number except 324 is 10

8)2nd term = 1st + 1 , 3rd term = 2nd +2 .........

9.> 66,36,18,?
a)3 b)6 c)8 d)12

10.>2,3,8,63,?
a)1038 b)1998 c)3008 d)3968

9. answer is 8
as
from 66 we have 6 x 6 = 36
from 36 we have 3 x 6 = 18
and finally from 18 we have 1 x 8 = 8

10.
it's like 2²-1 = 3
3² - 1 = 8
8² - 1 = 63
63² - 1 = 3968

10) 3968