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Problem C. 1143. (November 2012)

C. 1143. The first few terms of a number sequence are the even numbers 2, 4, 6, \ldots etc. From a certain term onwards, the sequence continues as an arithmetic progression with a common difference of d=3. Which term is that if the sum of the first 50 terms of the sequence is 2985?

(5 pont)

Deadline expired on December 10, 2012.

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

Megoldás. Mivel az első 50 elem összege páratlan szám, ezért biztos, hogy legkésőbb az 50. tagnál már 3 a differencia. Vagyis a sorozat így néz ki: \(\displaystyle 2, 4, 6, \ldots,2a, 2a+3, 2a+6,\ldots,2a+(50-a)\cdot3\) (ahol \(\displaystyle a>0\), egész).

A feladat szövege alapján:

\(\displaystyle 2985=\frac{(2+2a)\cdot a}{2}+\frac{[(2a+3)+(2a+(50-a)\cdot3)]\cdot(50-a)}{2}=\)

\(\displaystyle =\frac12(2a^2+2a+(2a+3+2a+150-3a)(50-a))=\)

\(\displaystyle =\frac12(2a^2+2a+(a+153)(50-a))=\frac12(2a^2+2a+7650-103a-a^2).\)


\(\displaystyle 5970=a^2-101a+7650,\)

\(\displaystyle 0=a^2-101a+1680.\)

Ebből \(\displaystyle a=\frac{101\pm\sqrt{3481}}{2}=\frac{101\pm59}{2}\). Vagyis \(\displaystyle a_1=80>50\), nem megoldás; \(\displaystyle a_2=21\).

Ekkor a keresett tag: \(\displaystyle 2a+3=45\).


306 students sent a solution.
5 points:151 students.
4 points:87 students.
3 points:39 students.
2 points:19 students.
1 point:5 students.
0 point:2 students.
Unfair, not evaluated:3 solutions.

Problems in Mathematics of KöMaL, November 2012