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Arithmetic Progression-2

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Carl Friedrich Gauss was in Grade 3 when his teacher was probably not in a mood to teachUntitled and told his students to find the sum of first 100 natural numbers. Within few minutes Gauss solved it. Here is a method how he solved the problem:

Gauss folded these numbers and rearranged them as shown. He knew that that the sum of these two sequences will give him the required result. He found that sum of all the vertical terms written is 101 (which is sum of first and the last term). Since there are 50 terms hence the sum of first 100 natural numbers would be 50 times 101 which is 5050.

 

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Around 500 A.D. Aryabhatta also used the similar method to find the sum of n terms of an AP.

Derivation to find Sum of first n terms of an AP.

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We have the following two formulas to find the sum of n terms of any arithmetic progression:
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If nth term of any sequence is given, then it forms an AP iff:

  1. It is quadratic in n.
  2. There is no constant term in the expression

moreover if we put n=1, we will get the firs term of the AP and if we double the coefficient of n^2 we will get the common difference and hence we can form an AP if its nth term is given.

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