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

  1. $1$-st
  2. $9$-th
  3. $12$-th
  4. None of these
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N th no. Of AP series= a+(N-1)d

a first element and d difference

Frm ques 3rd element a+2d

Similarly

 (a+2d) + (a+14d) = ( a+5d) + (a+10d) + (a+12d)

 => a+11d = 0

Its 12th element which is zero
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A common arithmetic progression is

a, a+d, a+2d, a+3d, a+4d,.....

where 'a' is the first element and 'd' is the common difference

As per the question,

the sum of 3rd and 15th elements is equal to the sum of 6th, eleventh and 13th elements, deriving from the formula

(a+2d) + (a+14d)=(a+5d)+(a+10d)+(a+12d)

2a+16d=3a+27d

2a+16d-3a-27d=0

a+11d=0

So the 12th element of the series is ZERO
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