Zad 1. Znajdz wzór ciągu arytmetycznego spełniającego podane warunki. Oblicz szósty wyraz tego ciągu.... Question from @Riff12

Riff12 @Riff12

October 2018 1 49 Report

Zad 1. Znajdz wzór ciągu arytmetycznego spełniającego podane warunki. Oblicz szósty wyraz tego ciągu.

a) a1 = 5 a3 = 8

b) a3 = 45 a7 = 34

c) a11= 0 a20 = -0,9

d) a5 = 31 a9 = 30

e) a4 = -10 a10 = 110

f) a11 = 4 a16 = 5

Z góry dziękuję.

wodnick

$a_{n}=a_{1}+(n-1)r\\\\ a_{1}=5, \ \ \ \ a_{3}=8\\\\ a_{3}=a_{1}+2r \ \ \Rightarrow \ \ a_{3}-a_{1}=2r \ \ \Rightarrow \ \ 8-5=2r \ \ \Rightarrow \ \ r=\frac{3}{2}\\\\ a_{n}=5+(n-1)*\frac{3}{2}=\frac{10+3n-3}{2}=\frac{3n+7}{2}$

$a_{3}=45, \ \ \ \ a_{7}=34\\\\ a_{7}=a_{3}+4r \ \ \Rightarrow \ \ a_{7}-a_{3}=4r \ \ \Rightarrow \ \ 34-45=4r \ \ \Rightarrow \ \ r=-\frac{11}{4}\\\\ a_{3}=a_{1}+2r \ \ \Rightarrow \ \ a_{1}=a_{3}-2r=45-2*(-\frac{11}{4})=\\=45+5\frac{1}{2}=50\frac{1}{2}\\\\ a_{n}=50\frac{1}{2}+(n-1)(-\frac{11}{4})=\frac{202}{4}-\frac{11n}{4}+\frac{11}{4}=\frac{213-11n}{4}$

$a_{11}=0, \ \ \ a_{20}=-0,9\\\\ a_{20}=a_{11}+9r \ \ \Rightarrow \ \ a_{20}-a_{11}=9r \ \ \Rightarrow \ \ -0,9-0=9r\\ \Rightarrow \ \ r=-\frac{0,9}{9}=-0,1\\\\ a_{11}=a_{1}+10r \ \ \Rightarrow \ \ a_{1}=a_{11}-10r=0-10*(-0,1)=10\\\\ a_{n}=10+(n-1)*(-0,1)=10-0,1n+0,1=10,1-0,1n$

$a_{5}=31, \ \ \ a_{9}=30\\\\ a_{9}=a_{5}+4r \ \ \Rightarrow \ \ a_{9}-a_{5}=4r \ \ \Rightarrow \ \ 30-31=4r \ \ \Rightarrow \ \ r=-\frac{1}{4}\\\\ a_{5}=a_{1}+4r \ \ \Rightarrow \ \ a_{1}=a_{5}-4r=31-4*(-\frac{1}{4})=31+1=32\\\\ a_{n}=32+(n-1)*(-\frac{1}{4})=32-\frac{1}{4}n+\frac{1}{4}=-\frac{1}{4}n+32\frac{1}{4}$

$a_{4}=-10, \ \ \ \ a_{10}=110\\\\ a_{10}=a_{4}+6r \ \ \Rightarrow \ \ a_{10}-a_{4}=6r \ \ \Rightarrow \ \ 110-(-10)=6r \ \ \Rightarrow \ \ \\ \Rightarrow \ \ r=20\\\\ a_{4}=a_{1}+3r \ \ \Rightarrow \ \ a_{1}=a_{4}-3r=-10-3*20=-10-60=-70\\\\ a_{n}=-70+(n-1)*20=-70+20n-20=20n-90$

$a_{11}=4, \ \ \ \ a_{16}=5\\\\ a_{16}=a_{11}+5r \ \ \Rightarrow \ \ a_{16}-a_{11}=5r \ \ \Rightarrow \ \ 5-4=5r \ \ \Rightarrow \ \ r=\frac{1}{5}\\\\ a_{11}=a_{1}+10r \ \ \Rightarrow \ \ a_{1}=a_{11}-10r=4-10*\frac{1}{5}=4-2=2\\\\ a_{n}=2+(n-1)*\frac{1}{5}=2+\frac{1}{5}n-\frac{1}{5}=\frac{1}{5}n+1\frac{4}{5}$