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Algebra Level 4

Let ∑ n \sum_{n} denote the sum of the first n n terms of a geometric progression.

Consider that ∑ 250 = 50 \sum_{250} = 50 and ∑ 750 = 650 \sum_{750} = 650 What is the greatest possible integer value for ∑ 500 \sum_{500} ?


The answer is 200.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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1 solution

We have that ∑ n ∑ 2 n − ∑ n = ∑ 2 n − ∑ n ∑ 3 n − ∑ 2 n \dfrac{\sum_{n}}{\sum_{2n} - \sum_{n}} = \dfrac{\sum_{2n} - \sum_{n}}{\sum_{3n} - \sum_{2n}} . (proof left to the reader).

Substituting, we find 50 x − 50 = x − 50 650 − x ⇔ x 2 − 50 x − 30000 = 0 → x 1 = − 150 , x 2 = 200. \frac{50}{x-50} = \frac{x-50}{650-x} \Leftrightarrow x^2 - 50x - 30000 = 0 \rightarrow x_1 = -150, \boxed{x_2 = 200.}

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