# 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).

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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

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