Which term of the series 5, 8, 11, 14,.....is 320?

A. 104th B. 105th
C. 106th D. 64th

• its an A.P. series
  an=a+(n-1)d where an(last term)=320;a(first term)=5;d(diff of two nos,)=3
  320=5+(n-1)3
  315/3= n-1
  n=106
  ans is c part
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• its ap series..
  l=a+(n-1)d
  l=last term
  n=nth
  d=diff
  320=5+(n-1)3
  n=106
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• it is an A.P
  Tn = a+(n-1)d
  => 320 = 5 + (n-1)3
  => 315 = (n-1)3
  => (n-1) = 105
  => n=106
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• 320 = 5 + (n-1)3
  315 = 3(n-1)
  n = 106
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• 320=5+(n-1)*3
  n=106
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• Ans : option C
  
  This can be found using arithmetic progression. In this first term is a,(a+d), (a+2d), (a+3d) and so on it goes.
  From the question we get,
  a = 5 ---------first term,
  (a+d) = (5 + 3 = 8) ----------second term,
  (a + 2d) = ( 5 + (2*3)) = 11 ------------third term
  
  therefore 106th term is ( a + 105d) = ( 5 + (105*3)) = 320
  
  ans : 106th term is 320
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