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Algebra Difficulty 3.1 AMC 10/12 Find the answer

If the sum of the first 3n3n positive integers is 150150 more than the sum of the first nn positive integers, then the sum of the first 4n4n positive integers is

Pick one

Solution

We can setup our first equation as
3n(3n+1)2=n(n+1)2+150\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150
Simplifying we get
9n2+3n=n2+n+300⇒8n2+2n−300=0⇒4n2+n−150=09n^2 + 3n = n^2 + n + 300 \Rightarrow 8n^2 + 2n - 300 = 0 \Rightarrow 4n^2 + n - 150 = 0
So our roots using the quadratic formula are
−b±b2−4ac2a⇒−1±12−4⋅(−150)⋅42⋅4⇒−1±1+24008⇒6,−25/4\dfrac{-b\pm\sqrt{b^2 - 4ac}}{2a} \Rightarrow \dfrac{-1\pm\sqrt{1^2 - 4\cdot(-150)\cdot4}}{2\cdot4} \Rightarrow \dfrac{-1\pm\sqrt{1+2400}}{8} \Rightarrow 6, -25/4
Since the question said positive integers, n=6n = 6, so 4n=244n = 24
24⋅252=300\frac{24\cdot 25}{2} = 300
A\fbox{A}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.