Summing AP again with 2

Algebra Level 3

Let a n a_n define a finite arithmetic progression such such that a 1 < 0 a_1 < 0 and let there exist a positive integer k ≠ 1 k \neq 1 such that a k = 0 a_k = 0 .

Find S 2 k − 1 \large S_{2k-1} , the sum to 2 k − 1 2k-1 terms of this arithmetic progression.


The answer is 0.000.

Viki Zeta by Viki Zeta , 19, India

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1 solution

Viki Zeta
Nov 2, 2016

S 2 k − 1 = 2 k − 1 2 ( 2 a + ( 2 k − 1 − 1 ) d ) = 2 k − 1 2 ( 2 a + ( 2 k − 2 ) d ) = 2 k − 1 2 2 ( a + ( k − 1 ) d ) = ( 2 k − 1 ) ( a + ( k − 1 ) d ) a k = 0 a + ( k − 1 ) d = 0 S 2 k − 1 = ( 2 k − 1 ) ( 0 ) ∴ S 2 k − 1 = 0 S_{2k-1} = \dfrac{2k-1}{2} \left(2a + (2k-1-1)d\right) \\ = \dfrac{2k-1}{2} \left(2a + (2k-2)d\right) \\ = \dfrac{2k-1}{2} 2\left(a + (k-1)d\right) \\ = (2k-1)\color{#D61F06}{(a + (k-1)d)} \\ a_k = 0 \\ \color{#D61F06}{a + (k-1)d} = \color{#3D99F6}{0} \\ S_{2k-1} = (2k-1)\color{#3D99F6}{(0)} \\ \boxed{\therefore S_{2k-1} = 0}

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Hey is your RMO out?

Md Zuhair - 4 years, 6 months ago

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no. not yet.

Viki Zeta - 4 years, 6 months ago

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Oho.... Our scores are out

Md Zuhair - 4 years, 6 months ago

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@Md Zuhair – website?.... oO

Viki Zeta - 4 years, 6 months ago

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@Viki Zeta – No.. our scores in WB region was emailed to us

Md Zuhair - 4 years, 6 months ago

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@Md Zuhair – oh... lol...

Viki Zeta - 4 years, 6 months ago

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