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Fie an progresie aritmetica:
Daca a1+a4=100
a indice(n-3)+an=200
Sn=600


Sa se determine n=?


Răspuns :

a₁+a₄=100 <=> a₁+a₁+3r=100 <=> 2a₁+3r=100 .

a de n +a de n-3 =200 <=> a₁+(n-1)·r +a₁+(n-4)·r=200 <=> 2a₁+(2n-5)·r=200 .

(2n-5-3)·r=100 <=> (2n-8)·r=100 <=> (n-4)·r=50 .

S de n=(a₁+a de n)·n /2=600 <=> (a₁+a de n)·n=1200 <=> (a₁+a₁+(n-1)·r)·n=1200 <=> (2a₁+(n-1)·r)·n=1200 => 150·n=1200 => n=8 .