find x in the series
3, 15, x, 51, 53,159,161

• answer is 17
  3,(3*5),(3*5+2)=17,(17*3=51),(51+2=53),(53*3=159),(159+2=161)
  
• 13 years agoHelpfull: Yes(20) No(7)
• u all have done (3*5) in the first but after the series goes with (17*3),(53*3)
  so i think if the first term will be 5 then in that case we can proceed with all of urs solution
• 13 years agoHelpfull: Yes(15) No(2)
• From the difference between 161 & 159 is 2. similarly between 51& 53 i 2.so the next sequence after 15 is added by 2. so answer is 17
• 13 years agoHelpfull: Yes(15) No(4)
• in every alternate term we add 2
  so 3,3*5=`15,15+2=17,17*3=51,51+2=53,53*3=159,159+2=161
  so answer is 17
• 13 years agoHelpfull: Yes(6) No(2)
• answer is 17
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• the answer is 17 because 3*5=15 then 15+2=17 then 17*3=51 again 51+2=53 then 53*3=159 so this series is continuous........so answer is 17
  
• 9 years agoHelpfull: Yes(0) No(0)
• 17
  15+2=17
  17*3=51
• 7 years agoHelpfull: Yes(0) No(0)
• 3(1st term)
  3*5=15(2nd Term)
  3*5+2=17(3rd Term)
  17*3=51(4th term)
  17*3+2=53(5th term)
  53*3=159(6th term)
  53*3+2=161(7th term)
• 6 years agoHelpfull: Yes(0) No(0)