Problem C. 1087. (September 2011)

C. 1087. The first term of an arithmetic progression is 1, its second term is n, and the sum of the first n terms is 33n. Find n.

(5 pont)

Deadline expired on October 10, 2011.

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

Megoldás. Tudjuk, hogy \(\displaystyle a_1=1\) és \(\displaystyle a_2=n\), ebből \(\displaystyle d=a_2-a_1=n-1\).

Mivel ez egy számtani sorozat, így

\(\displaystyle a_n=a_1+(n-1)\cdot d=a_1+(n-1)^2.\)

Mivel

\(\displaystyle 33n=S_n=\frac {a_1+a_n} {2} \cdot n=\frac {a_1+a_1+(n-1)^2} {2} \cdot n,\)

így

\(\displaystyle 33n=\frac {1+1+(n-1)^2} {2} \cdot n,\)

\(\displaystyle 33=\frac {2+(n-1)^2} {2},\)

\(\displaystyle 66=2+(n-1)^2,\)

\(\displaystyle 64=(n-1)^2.\)

Ebből \(\displaystyle n-1=\pm8\). Mivel \(\displaystyle n\) pozitív, ezért csak \(\displaystyle n-1=8\) lehet megoldás, ekkor \(\displaystyle n=9\).

Ez elkenőrizve teljesíti a feltételeket: \(\displaystyle a_1=1\) és \(\displaystyle a_2=9\) esetén \(\displaystyle d=8\) és \(\displaystyle S_9=33\cdot9\).

Statistics:

481 students sent a solution.
5 points:	357 students.
4 points:	85 students.
3 points:	12 students.
2 points:	9 students.
1 point:	6 students.
0 point:	7 students.
Unfair, not evaluated:	5 solutionss.

Problems in Mathematics of KöMaL, September 2011