Math, asked by aachman68, 10 months ago

Let \displaystyle {a_n}an be a finite arithmetic progression and k be a natural number. \displaystyle a_1=r < 0a1=r<0 and \displaystyle a_k=0ak=0. Find \displaystyle S_{2k-1}S2k−1 (the sum of the first 2k-1 elements of the progression).

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Answered by vaibhavi7742

The sum of n terms of AP is the sum(addition) of first n terms of the arithmetic sequence. It is equal to n divided by 2 times the sum of twice the first term – 'a' and the product of the difference between second and first term-'d' also know as common difference, and (n-1), where n is numbers of terms to be added.

Answered by nu1273311

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Step-by-step explanation:

It's n easy questions

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Let \displaystyle {a_n}an​ be a finite arithmetic progression and k be a natural number. \displaystyle a_1=r < 0a1​=r<0 and \displaystyle a_k=0ak​=0. Find \displaystyle S_{2k-1}S2k−1​ (the sum of the first 2k-1 elements of the progression). ​

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Let \displaystyle {a_n}an be a finite arithmetic progression and k be a natural number. \displaystyle a_1=r < 0a1=r<0 and \displaystyle a_k=0ak=0. Find \displaystyle S_{2k-1}S2k−1 (the sum of the first 2k-1 elements of the progression).