Mathematical and Physical Journal
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Problem C. 1154. (January 2013)

C. 1154. Is there an arithmetic progression in which the sum of the first n terms is n, the sum of the first 2n terms is n2, and the sum of the first 3n terms is n3?

(5 pont)

Deadline expired on February 11, 2013.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A sorozat első elemét jelölje \(\displaystyle a_1\), a differenciát pedig \(\displaystyle d\). Ekkor:

\(\displaystyle s_n=\frac{2a_1+(n-1)d}{2}\cdot n=n, \)

\(\displaystyle s_{2n}=\frac{2a_1+(2n-1)d}{2}\cdot 2n=n^2, \)

\(\displaystyle s_{3n}=\frac{2a_1+(3n-1)d}{2}\cdot 3n=n^3. \)

Az első egyenletből:

(1)\(\displaystyle 1=a_1+\frac{(n-1)d}{2},\)

a másodikból:

(2)\(\displaystyle \frac n2=a_1+\frac{(2n-1)d}{2},\)

a harmadikból:

(3)\(\displaystyle \frac {n^2}{3}=a_1+\frac{(3n-1)d}{2}.\)

2)-ből kivonva 1)-et:

\(\displaystyle \frac n2-1=\frac n2\cdot d,\)

\(\displaystyle n-2=nd.\)

3)-ból 1)-et kivonva:

\(\displaystyle \frac{n^2}{3}-1=nd.\)

A két kapott egyenlet jobb bal oldala egyenlő, így a bal oldaluk is az:

\(\displaystyle \frac{n^2}{3}-1=n-2,\)

\(\displaystyle \frac{n^2}{3}-n+1=0,\)

\(\displaystyle n^2-3n+3=0.\)

Ennek az egyenletnek a diszkriminánsa negatív, így nincs megoldása.


Statistics:

202 students sent a solution.
5 points:127 students.
4 points:13 students.
3 points:18 students.
2 points:7 students.
1 point:10 students.
0 point:27 students.

Problems in Mathematics of KöMaL, January 2013