Question

72 W ciągu arytmetycznym (an) suma wyrazów o numerach 17 i 81 jest równa ‒282, a suma wyrazów o numerach 42. i 68. wynosi ‒322. Oblicz a99 + a47.

Answer 1

Odpowiedź:

[tex]a_{99}+a_{47}=-442[/tex]

Szczegółowe wyjaśnienie:

[tex]a_{17}+a_{81}=-282\\a_{42}+a_{68}=-322\\a_{99}+a_{47}=?\\\\a_{17}=a_1+16r\\a_{81}=a_1+80r\\a_{42}=a_1+41r\\a_{68}=a_1+67r\\a_{99}=a_1+98r\\a_{47}=a_1+46r[/tex]

[tex]\left \{ {{a_{17}+a_{81}=-282} \atop {a_{42}+a_{68}=-322}} \right. \\\left \{ {{a_1+16r+a_1+80r=-282} \atop {a_1+41r+a_1+67r=-322}} \right. \\\left \{ {{2a_1+96r=-282\ |:(-2)} \atop {2a_1+108r=-322\ |:2}} \right. \\\left \{ {{-a_1-48r=141} \atop {a_1+54r=-161}} \right|+ \\\left \{ {{6r=-20\ |:6} \atop {a_1+54r=-161}} \right. \\\left \{ {{r=-3\frac{1}{3}} \atop {a_1+54*\left(-3\frac{1}{3}\right)=-161}} \right. \\\left \{ {{r=-3\frac{1}{3}} \atop {a_1+54*\left(-\frac{10}{3}\right)=-161}} \right.[/tex]

[tex]\left \{ {{r=-3\frac{1}{3}} \atop {a_1-180=-161}} \right. \\\left \{ {{r=-3\frac{1}{3}} \atop {a_1=-161+180}} \right. \\\left \{ {{r=-3\frac{1}{3}} \atop {a_1=19}} \right.[/tex]

Zatem

[tex]a_{99}+a_{47}=a_1+98r+a_1+46r=2a_1+144r=2*19+144*\left(-3\frac{1}{3}\right)=\\=38+144*\left(-\frac{10}{3}\right)=38+48*\left(-\frac{10}{1}\right)=38-480=-442[/tex]