# Let be a finite arithmetic progression and k be a natural number. and . Find (the sum of the first 2k-1 elements of the progression).

1
by ankitkumar0102

2015-04-05T16:53:33+05:30

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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
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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