Aflati primul termen si ratia unei progresii aritmetice stiind ca S11=143 si a2a4=7
albatran:
a2-a4...
Răspunsuri la întrebare
Răspuns de
0
a2-a4=2r=7⇒r=7/2
S11=a1+a1+r+a1+2r+...+a1+10r=11a1+(10*11/2) *r=11a1+55r
143=11a1+55r
13=a1+5r
a1=13-5r
a1=13-5*7/2= 13-35/2=(26-35)/2=-9/2
Verificare;
an= -9/2;-1;5/2;6;19/2........28; 61/2
S11= 11/2*(61/2-9/2)=(11/2)*26=11*13=143
adevarat, problema este bine rezolvata
S11=a1+a1+r+a1+2r+...+a1+10r=11a1+(10*11/2) *r=11a1+55r
143=11a1+55r
13=a1+5r
a1=13-5r
a1=13-5*7/2= 13-35/2=(26-35)/2=-9/2
Verificare;
an= -9/2;-1;5/2;6;19/2........28; 61/2
S11= 11/2*(61/2-9/2)=(11/2)*26=11*13=143
adevarat, problema este bine rezolvata
Alte întrebări interesante