Q. The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

Option
a) 1
b) 9
c) 12
d) None

• 12
  
  If we consider the third term to be ‘x”
  The 15th term will be (x + 12d)
  6th term will be (x + 3d)
  11th term will be (x + 8d) and
  13th term will be (x + 10d).
  Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.
  x + 9d will be the 12th term.
  
• 13 years agoHelpfull: Yes(21) No(0)
• first term=a+0d
  2nd term=a+d;
  lly 3rd term=a+2d
  
  so according to question:
  a+2d+a+14d=a+5d+a+10d+a+11d
  2a+16d=3a+26d
  a+11d=0
  i.e 12 term = zero...
• 9 years agoHelpfull: Yes(6) No(0)
• 12
  
  Really nice solution given by Dipin.
• 13 years agoHelpfull: Yes(2) No(2)
• 12: as as by solving that we get a+11d=0,this implies that 12th term =zero
  
• 9 years agoHelpfull: Yes(0) No(0)
• t3 + t5 = t6+t11+t13
  
  t3 = a + 2d
  => a+2d + a +14d = a +5d + a +10d + a + 12d
  => a + 11d = 0
  => 12th term gives 0 .(Ans)
• 9 years agoHelpfull: Yes(0) No(0)
• Given==> t3 + t5 = t6+t11+t13
  
  We know===> t3 = a + 2d
  =>( a+2d )+ (a +14d) = (a +5d) +( a +10d) + (a + 12d)
  =>2a+16d = 3a+27d
  => a + 11d = 0
  => 12th term gives 0
  Therefore option "C" is the answer
• 6 years agoHelpfull: Yes(0) No(0)