Question

fie (an)n>=1 o progresie aritmetica.
Determinati n€N* pentru care a1+a4=100 , an-3+an=200 si Sn=600​

Answer 1

Răspuns:

$a_{1}+a_{4}=100,~deci~a_{1}+a_{1}+3r=100,~deci~2a_{1}+3r=100.~(*)\\a_{n-3}+a_{n}=200,~Scadem~primele~doua~egalitati~date,~parte~cu~parte\\(a_{n-3}+a_{n})-(a_{1}+a_{4})=200-100,~(a_{n-3}-a_{1})+(a_{n}-a_{4})=100,\\(n-3-1)r+(n-4)r=100,~2*(n-4)r=100,~(n-4)r=50.~(**)\\S_{n}=\frac{n(2a_{1}+(n-1)r)}{2} =~ |aplicam~ (*)~|~=\frac{n(100-3r+(n-1)r)}{2} \\aplicam~(**),~(n-4)r=50,~(n-1-3)r=50,~(n-1)r-3r=50,~(n-1)r=3r+50.~inlocuim~la~suma~S_{n}=\frac{n(100-3r+(n-1)r)}{2}=\frac{n(100-3r+3r+50)}{2}=n*75=600,$

deci n=600:75=8

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