Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# 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 solutions.

Problems in Mathematics of KöMaL, September 2011