1970 AMC 12 Problem 22

Attempt Problem 22 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

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22.

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:

300

350

400

450

600

Use

1+2+\cdots+m=\frac{m(m+1)}{2}

The given difference simplifies to

4n^2+n

Let

S_m=\frac{m(m+1)}{2}.

Then

\begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n. \end{aligned}

Since

S_{3n}-S_n=150,

this gives

4n^2+n=150.

Hence

(n-6)(4n+25)=0,

and the positive solution is

n=6.

Therefore

S_{4n}=S_{24}=\frac{24\cdot25}{2}=300.

Therefore, the correct answer is A.

Problem 22 in other years