Arithmetic Series Questions (Problem 2)

Number Theory Level pending

Using the formula S n S_n = n 2 \frac{n}{2} [ 2 a + ( n − 1 ) d ] [2a + (n - 1)d] , find the sum of the arithmetic series that starts from 1 and finishes at -500,000 (1 + 0 + ... + -500,000).


The answer is 124999750000.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Using S n S_n = n 2 \frac{n}{2} [ 2 a + ( n − 1 ) ∗ d ] [2a + (n - 1) * d] : a = 1, n = -500,000 and d = 1 so S n S_n = − 500 , 000 2 \frac{-500,000}{2} [ 2 ∗ 1 + ( − 500 , 000 − 1 ) ∗ 1 ] [2 * 1 + (-500,000 - 1) * 1] = -250,000 [ 2 ∗ 1 + ( − 500 , 000 − 1 ) ∗ 1 ] [2 * 1 + (-500,000 - 1) * 1] = -250,000 [ 2 + ( − 500 , 001 ) ∗ 1 ] [2 + (-500,001) * 1] = -250,000 [ 2 + ( − 500 , 001 ) ] [2 + (-500,001)] = -250,000 [ ( − 499 , 999 ) ] [(-499,999)] = -250,000(-499,999) = 124,999,750,000

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How can n be negative? n is known as the number of terms which must be positive! Also, d=-1.

Isaac YIU Math Studio - 1 year, 1 month ago

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I responded to your report - can you please give me the working out I asked for so I can rectify it?

A Former Brilliant Member - 1 year, 1 month ago

I ask the Brilliant staff to rectify the problem if it's wrong.

A Former Brilliant Member - 1 year, 1 month ago

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