Question

Scrieti primii cinci termeni ai progresiei aritmetice in cazurile:
1)a1=1/3, r=1; 2)a1=-1, r=-2; 3)a1=-2, r=3; 4)a1=-0,5, r=1/2; 5)a1=1/3, r=1/2.
Va rog raspundeti repede. Multumesc.

Answer 1

$\boxed{a_n=a_{n-1}+r}$

$\displaystyle 1).a_1= \frac{1}{3} ~,~r=1 \\ \\ a_2=a_{2-1}+r=a_1+r= \frac{1}{3} +1= \frac{1+3}{3} = \frac{4}{3} \\ \\ a_3=a_{3-1}+r=a_2+r= \frac{4}{3} +1= \frac{4+3}{3} = \frac{7}{3} \\ \\ a_4=a_{4-1}+r=a_3+r= \frac{7}{3} +1= \frac{7+3}{3} = \frac{10}{3} \\ \\ a_5=a_{5-1}+r=a_4+r= \frac{10}{3} +1= \frac{10+3}{3} = \frac{13}{3} \\ \\ \frac{1}{3} , \frac{4}{3} , \frac{7}{3} , \frac{10}{3} , \frac{13}{3} ,...$

$\displaystyle 2).a_1=-1~,~r=-2 \\ \\ a_2=a_{2-1}+r=a_1+r=-1+(-2)=-1-2=-3 \\ \\ a_3=a_{3-1}+r=a_2+r=-3+(-2)=-3-2=-5 \\ \\ a_4=a_{4-1}+r=a_3+r=-5+(-2)=-5-2=-7 \\ \\ a_5=a_{5-1}+r=a_4+r=-7+(-2)=-7-2=-9 \\ \\ -1,-3,-5,-7,-9,...$

$\displaystyle 3).a_1=-2~,~r=3 \\ \\ a_2=a_{2-1}+r=a_1+r=-2+3=1 \\ \\ a_3=a_{3-1}+r=a_2+r=1+3=4 \\ \\ a_4=a_{4-1}+r=a_3+r=4+3=7 \\ \\ a_5=a_{5-1}+r=a_4+r=7+3=10 \\ \\ -2,1,4,7,10,...$

$\displaystyle 4).a_1=-0,5=- \frac{5}{10} \Rightarrow a_1=- \frac{1}{2} ~,~r= \frac{1}{2} \\ \\ a_2= a_{2-1}+r=a_1+r=- \frac{1}{2} + \frac{1}{2} =0 \\ \\ a_3= a_{3-1}+r=a_2+r=0+ \frac{1}{2} = \frac{1}{2} \\ \\ a_4=a_{4-1}+r=a_3+r= \frac{1}{2} + \frac{1}{2}= \frac{2}{2} =1 \\ \\ a_5=a_{5-1}+r=a_4+r=1+ \frac{1}{2} = \frac{2+1}{2}= \frac{3}{2} \\ \\ - \frac{1}{2} ,0, \frac{1}{2} ,1, \frac{3}{2},...$

$\displaystyle 5).a_1= \frac{1}{3} ~,~r= \frac{1}{2} \\ \\ a_2=a_{2-1}+r=a_1+r= \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \\ \\ a_3=a_{3-1}+r=a_2+r= \frac{5}{6} + \frac{1}{2}= \frac{5}{6}+ \frac{3}{6} = \frac{8}{6} = \frac{4}{3} \\ \\ a_4=a_{4-1}+r=a_3+r= \frac{4}{3} + \frac{1}{2} = \frac{8}{6} + \frac{3}{6}= \frac{11}{6} \\ \\ a_5=a_{5-1}+r=a_4+r= \frac{11}{6} + \frac{1}{2} = \frac{11}{6} + \frac{3}{6} = \frac{14}{6}= \frac{7}{3}$

$\displaystyle \frac{1}{3} , \frac{5}{6} , \frac{4}{3} , \frac{11}{6} , \frac{7}{3} ,...$