87th number in the series 2, 10, 26,
50,……….

• 2+8*0=2
  2+8*1=10
  2+8*1+8*2=26
  2+8*1+8*2+8*3=50...........
  87th term
  2+8*1+8*2+8*3+.......................................+8*86
  2+8(1+2+3+4+....................+86)
  2+8*(86/2)*(1+86)= 29930
• 9 years agoHelpfull: Yes(70) No(2)
• how can we take d=8. ? its not an AP
• 9 years agoHelpfull: Yes(4) No(4)
• here we can use formula : an=a+(n-1)*d
  a=2
  n=87
  d=8
  an=2+86*8
  an=690
• 9 years agoHelpfull: Yes(3) No(32)
• 2
  2+8
  2+8+8*2
  2+8+8*2+8*3
  87th term= 2+ 8*(1+2+3+.....86)=2+8*86/2*87=2+4*86*87=29930
• 9 years agoHelpfull: Yes(3) No(1)
• 1^2+1,3^2+1,5^2+1................
  A+(N-1)D=1+(87-1)*2=29929
  29929+1=29930
• 9 years agoHelpfull: Yes(3) No(2)
• diff b/w 2 and 10 is 8
  diff b/w 10 and 26 is 16
  diff b/w 26 and 50 is 24
  so diff series is 8, 16 , 24,and so on series continues..... so find 86th term in this series i.e.29928 and add 2 to that, then we will get actual 87th term in the required series that is 29930.
  
• 9 years agoHelpfull: Yes(2) No(0)
• 87th no. is 700.
  2nd term=1st term+8
  3rd term=1st term+2(8)
  .
  .
  87th term=1st term+86(8)
  =2+688=700
• 9 years agoHelpfull: Yes(2) No(3)
• mandira your answer is correct...kshitij and sitanshu u both are finding the sum upto 87th no..we just have to find 87th no..
  if we subtract second no by first we get an secondary AP..
  8,16,24...
  
  now take a=8 n=86 d=8
  
  t87=8+(86-1)*8
  t87=688
• 9 years agoHelpfull: Yes(1) No(11)
• since 87th number is required:
  86*8= 688
• 9 years agoHelpfull: Yes(0) No(5)
• And is 82
  10-2=8 =8*1
  26-10=16=8*2
  50-26=24=8*3
  So the next no. will be 50+(8*4)=50+32=82
• 9 years agoHelpfull: Yes(0) No(6)
• As the difference among the terms is constant.....apply Tn=an^2+bn+c.
  
• 9 years agoHelpfull: Yes(0) No(1)