Question

*Rezolvati :

1) Scrieti primii patru termeni ai unei progresii aritmetice daca :
a) a1 = -5 ; r = 4.
b) a1 = 1/2 ; r = 3.
c) a1 = -1/3 ; r = -1/2.

2) Scrieti termenii de rang 4;9;12 din ÷ (an) daca :
a) a2 = 5 ; r = -3.
b) a3 = 2 ; r = -2.
c) a6 = 8 ; r = 1/2.

3) Calculati :
a) 28+30+32+....+544.
b) 33+36+39+....+90.

Vreau rezolvarea completa va rog !

Answer 1

1)
a) a1 = -5 ; r = 4.
a1 = -5 +0×4= -5
a2= -5+1*4= -1
a3= -5+2*4=  3
a4=-5+3*4= 7

b)a1 = 1/2 ; r = 3.
a1= 1/2+0*3=1/2
a2=1/2+ 1*3=1/2+3=1/2+6/2=7/2
a3=1/2+2*3=1/2+6=1/2+12/2=13/2
a4=1/2+3*3=1/2+9=1/2+18/2=19/2


c) a1 = -1/3 ; r = -1/2.
a1= -1/3+ (-1/2)*0= -1/3
a2= -1/3+(-1/2)*1= -1/3-1/2=-2/6-3/6= -5/6
a3=-1/3+(-1/2)*2= - 1/3 -1= -4/3
a4= -1/3+ (-1/2)*3 = - 1/3-3/2= -2/6 -9/6= -11/6

2.
a) a2 = 5 ; r = -3.⇒ a1= 5+3=8; a2= 8+(-3)*0=8-3=5

a4=8+(-3)*3=8-9= -1
a9= 8+(-3)*8=8-24= -16
a12= 8+(-3)*11=8-33= -25

b) a3 = 2 ; r = -2.  ⇒ a2= 2+2=4; a1= 2+4=6

a4=6+(-2)*3=6-6=0
a9=6+(-2)*8=6- 16= -10
a12=6+(-2*11)=6-22= -16

c) a6 = 8 ; r = 1/2.
a4 = a6- 2*1/2=8-1=7
a9 =a6+1/2*3=8+3/2=19/2
a12= a6+1/2*6=8+3=11

3.
3) Calculati :
a) 28+30+32+....+544=(28+544)×n/2=572×259÷2=74074
a1=28; r=2; a2=28+2*1=30
an= 28+(n-1)*2
28+(n-1)*2=544
28+2n-2=544
2n=518
n=259

 b) 33+36+39+....+90= (33+90)×20÷2=123×10=1230
a1=33+3*0=33
an=33+(n-1)×3
33+(n-1)×3=90
33+3n-3=90
3n=60
n=20