. Inspired by Fibonacci series Sangeet decided to create his own series which is 1, 2, 3, 7, 7, 22, 15, 67, 31, _, 63?
a) 202 b) 31 c) 76 d) 49

• ans :a(202)
  consider alternative numbers 1,3,7,15,31,63
  1*2+1=3,3*2+1=7,7*2+1=15,31*2+1=63
  conside now remaing digits 2,7,22,67,-
  2*3+1=7,7*3+1=22,22*3+1=67,67*3+1=202
• 12 years agoHelpfull: Yes(47) No(3)
• Ans: 202
  its the alternative series connection
  series- 1,3,7,15,31,63
  as 1x2+1=3, 3x2+1=7, 7x2+1=15 and so on
  likewise other series- 2,7,22,67,?
  2x3+1=7, 7x3+1=22, 22x3+1=67, 67x3+1=202
• 12 years agoHelpfull: Yes(12) No(1)
• Ans=202
  bcos in the given series there are two other series..in the form of
  1 ,2  ,3  ,7 ,7  ,22  ,15 ,67  ,31  ,_ ,63;
  choose the alternate digits:1,3,7,15,31,63
  these are follow the pattern like 7=n*2+1(wher n is the previous number)
  15=7*2+1(where n=7)
  31=15*2+1(wher n=15)
  63=31*2+1(where n=31)
  but we dont need this series.just for understanding i gave the above matter
  anothe series in the given series in:2 ,7 ,22 ,67 ,_
  we hav to find the number in the blank
  this series is in the form of n*3+1
  so the series consists of  the following numbers  2
  7=2*3+1(where n=2)
  22=7*3+1(wher n=7)
  67=22*3+1(wher n=22)
  for the next number we need to multiply the previous number with 3 and add a 1 to the result
  finally next digit=67*3+1=202
  the final series 1,2,3,7,7,22,15,67,31,202,63
  
• 12 years agoHelpfull: Yes(4) No(3)
• ans : 202
  there are 2 series given
  1st : 1,3,7,15,31,63
  1+2 = 3+2(2) = 7+4(2) = 15+8(2) = 31+16(2) = 63
  2nd : 2,7,22,67,_
  2+5 = 7+5(3) = 22+5(3*3) = 67+5(3*3*3) = 202
• 8 years agoHelpfull: Yes(0) No(0)
• 31
  being repeated
  
• 7 years agoHelpfull: Yes(0) No(0)