1970 AHSME Problems/Problem 22

Problem

If the sum of the first $3n$ positive integers is $150$ more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is

$\text{(A) } 300\quad \text{(B) } 350\quad \text{(C) } 400\quad \text{(D) } 450\quad \text{(E) } 600$

Solution 1

We can setup our first equation as

$\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150$

Simplifying we get

$9n^2 + 3n = n^2 + n + 300 \Rightarrow 8n^2 + 2n - 300 = 0 \Rightarrow 4n^2 + n - 150 = 0$

So our roots using the quadratic formula are

$\dfrac{-b\pm\sqrt{b^2 - 4ac}}{2a} \Rightarrow \dfrac{-1\pm\sqrt{1^2 - 4\cdot(-150)\cdot4}}{2\cdot4} \Rightarrow \dfrac{-1\pm\sqrt{1+2400}}{8} \Rightarrow 6, -25/4$

Since the question said positive integers, $n = 6$ , so $4n = 24$

$\frac{24\cdot 25}{2} = 300$

$\fbox{A}$

Solution 2

Expressing as an equation: \begin{equation}\tag{1}\frac{3n(3n+1)}{2} = \frac{n(n+1)}{2} + 150. \end{equation}

The sum of the first 4n positive integers = \begin{equation}\tag{2}\frac{4n(4n+1)}{2}\end{equation}

We will try to rearrange Equation (1) to give equation (2)

$\frac{3n(3n+1)}{2} - \frac{n(n+1)}{2} = 150$

$=\frac{n(3(3n+1)-(n+1))}{2} = 150 =\frac{n(9n+3-n-1)}{2}$

$\frac{n(8n+2)}{2}= \frac{2n(4n+1)}{2} = 150$

$\frac{4n(4n+1)}{2} = 2*150 = 300$

300 is the answer

$\fbox{A}$

〜Melkor

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1970 AHSME Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

See also