AB EF JK QR ?
a)YZ
b)ZA
c)AB
d)None

• b)ZA
  
  In the given series,
  
  Intervals between every set of letters is in consecutive ODD numbers.
  
  Interval between AB and EF = 2 letters (C and D)
  Interval between EF and JK = 3 letters (G,H and I)
  Interval between JK and QR = 5 letters (L,M,N,O and P)
  So, next interval is 7 letters.
  Therefore,
  Next set of letters = ZA
  
  Interval between QR and ZA = 7 letters (S,T,U,V,W,X and Y)
  
  Final Sequence = AB EF JK QR ZA
  
• 12 years agoHelpfull: Yes(32) No(18)
• characters left between ab and ef =2
  characters left between ef and jk=3
  haracters left between jk and qr=5
  2+3=5
  =>3+5=8
  so leave 8 characters so the next term will be ab..
• 12 years agoHelpfull: Yes(30) No(12)
• AB "cd" EF "ghi" JK "lmnop" QR "stuvwxy" ZA
  
  In the given sequence the missing letters form prime numbers 2,3,5 and therefore the last must be 7.
  
  The answer is (b)ZA..
• 12 years agoHelpfull: Yes(8) No(3)
• the ans is option b)ZA.
  gap between AB and EF is 2,EF and JK is 3,JK and QR is 5.
  so similarly next gap should be 7.
  so the ans is ZA.
• 12 years agoHelpfull: Yes(1) No(7)
• b) ZA. the gap betwen frst two sequence is 2 (i.e C & D)...the gap between 2nd and 3rd is 3 (i.e G,H and I)..the next sequence is obtained by the adding the previous two gaps (i.e 2+3=5) so we get QR...similarly the nxt one would be the result pf adding prev two gaps i.e 3+5=8 .....hence we get ZA...
• 12 years agoHelpfull: Yes(1) No(6)
• the ans is option b)ZA.
  gap between AB and EF is 2,EF and JK is 3,JK and QR is 5.
  so similarly next gap should be 7.
  so the ans is ZA.
• 12 years agoHelpfull: Yes(1) No(5)
• YZ is the answer because
  AB EF JK QR ZA
  CD-2 GHI-3 LMNOP-5 STUVWXY-7
  so all are prime numbers after QR 7th letter after R is ZA
• 12 years agoHelpfull: Yes(1) No(5)
• AB CD2 EF GHI3 JK LMNOP4 QR STUVWX5 YZ
• 12 years agoHelpfull: Yes(0) No(4)