Question

Se considera o progresie aritmetica astfel incat a3=5 si a5=11.Calculati suma primulor 7 termenii ai progresiei va rog.

Answer 1

3a=5
5a=11
1a+2a+3a+4a+5a+6a+7a=5a+2a=11+6=17
2a=5a-3a=11-5=6

Answer 2

[tex]a3=5 a5=11 r=a5-a3 2r=11-5 2r=6 r= 6/2 =3 [/tex]

[tex]a1+a2+a3+a4+a5+a6+a7 [/tex]

[tex]a3=a1+2r 5=a1+6 5-a1-6=0 -a1-1=0 -a1=1 /(-1) a1=-1 [/tex]


$a2=a1+r = -1+3=2$

[tex]a4=a1+3r= -1+9=8 a5=a1+4r=-1+12=11 a6=a1+5r=-1+15=14 a7=a1+6r=-1+18=17[/tex]


-1+2+5+8+11+14+17=56