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Algebra Level 2

Consider an arithmetic progression sum $1+2+3+\cdots$ .

1 is the first term, 2 is the second term, etc.

The cumulative sum is obtained by adding all the terms up to a certain point. For example, the cumulative sum of the first 5 terms is $1 + 2 + 3 + 4 + 5 = 15$ .

How many terms out does the cumulative sum first exceed 100?

The answer is 14.

2 solutions

Denton Young
May 29, 2016

The cumulative sum is $\dfrac{1}{2}$ * n * (n+1) for the first n terms.

For n = 13, this is 91. For n = 14, this is 105.

Moderator note:

Simple standard approach.

A quick approximation for the number to get to the target number is sqrt(2*(target number)) This can be derived from the above equation if we assume n(n+1) ~ n^2.

Alex Li - 4 years, 2 months ago

Hung Woei Neoh
May 29, 2016

This is an arithmetic progression where $a=1,\;d=1$

Now, we want the sum of the first $n$ terms to be over $100$ , which is:

$S_n > 100\\ \dfrac{n}{2}\left(2(1) + (n-1)(1)\right) > 100\\ n(2+n-1) > 200\\ n^2+n-200 > 0$

First, if you solve $n^2+n-200=0$ , you should get $x=\dfrac{-1 \pm \sqrt{801}}{2}$

Therefore, for the inequality $n^2+n-200=0$ , we have the range:

$n < \dfrac{-1-\sqrt{801}}{2}$ or $n > \dfrac{-1+\sqrt{801}}{2}$

Since $n$ can only take positive values, the range is

$n > \dfrac{-1+\sqrt{801}}{2}\\ n > 13.65$

Therefore, the minimum number of terms required is $\boxed{14}$

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The answer is 14.

2 solutions

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