NET JRF PART - A : Questions and Solutions : NET DEC-2014

NET JRF PART - A : Questions and Solutions:

NET DEC-2014

1.) Continue the sequence 

  2 , 5, 10, 17, 28, 41, _, _, _

(a) 58, 77, 100  (b) 64, 81, 100 (c) 43, 47, 53 (d) 55, 89, 113

Ans. : (a) 58, 77, 100 .

Solution: 

5 – 2 = 3, 10 – 5 = 5, 17 – 10 = 7, 28 – 17 = 11, 41 - 28 = 13 . 

Here we see that, difference between consecutive term is a prime number. Hence the next 

three numbers are obtained by adding 17, 19 and 23 to the preceding term respectively. 

 41 + 17 = 58,

 58 + 19 = 77,

 77 + 23 = 100 .

The sequence is 2 , 5, 10, 17, 28, 41, 58, 77, 100 .