Question

Let ${a_n}$ be a finite arithmetic progression and k be a natural number. $a_1=r < 0$ and $a_k=0$. Find $S_{2k-1}$ (the sum of the first 2k-1 elements of the progression).

Answer 1

a₁ = r  <  0            the first number of AP is less than zero.
   a_k =  0,      k > 0

It means that  common difference d  is positive.
 a_k = a₁ + (k-1) d  = 0
       = r + (k-1) d = 0
  =>   d = - r / (k -1)

 S_2k-1  = sum of 1st 2k-1 terms of the AP
          = [ 2 a₁ + (2k-2) d ] (2k-1)/2
          = [ 2 r  -  2 (k-1) r /(k-1) ] *(2k -1) /2
           = 0
============
it is seen that a_1 to  a_k-1  are negative.  a_k = 0.
         a_k+1 to  a_2k-1 are positive.    These k-1 terms are symmetrically equal to  a_1 to a_k-1 , but opposite in sign.
   Thus the sum  a_1 +.... + a_k-1 + a_k + a_k+1 + ..... + a_2k-1  =  a_k = 0