Question

Sa se determine progresiile geometrice in cazurile :
a) a1=3, Sn=765, q=2
b) a4= 135, a7=3645
c) a5=512, a7=8192

Answer 1

a)$S_{n} = a_{1}$·$\frac{1- q^{n} }{1-q}$=3·$\frac{1- 2^{n} }{1-2}$=765
1-$2^{n}$=-255
$2^{n}$=256=$2^{8}$⇒n=8
$a_{n} = a_{1}$·$q^{n-1}$⇒
a1=3; a2=3·2=6; a3=3·4=12....... a8=3·$2^{7}$=384
b) $a_{4} = a_{1}$·$q^{3}$⇒135=$a_{1}$·$q^{3}$
$a_{7} =3645= a_{1}$·$q^{6}$
impartim a7 la a4⇒$\frac{ a_{7} }{ a_{4} } = \frac{3645}{135} = \frac{ a_{1} q^{6} }{ a_{1} q^{3} }$
⇒$q^{3} =27$
q=3⇒a1=5; a2=5·3=15; a3=5·3²=45...........
c) se rezolva la fel ca b)
incearca sa il rezolvi si daca ai probleme ma intrebi